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THE 



ELEMENTS 



OF 



ANALYTICAL GEOMETRY. 



By ELIAS LOOMIS, LL.D., 

PROFESSOR OF NATURAL PHILOSOPHY AND ASTRONOMY IN YALE COLLEGE, 
AND AUTHOR OF " A COURSE OF MATHEMATICS." 



EEVISED EDITION. 





T/> 720? V^ 



NEW YORK: 

HARPER & BROTHERS, PUBLISHERS, 

FRANKLIN SQUARE. 

1873. 



A 

V 1 



Entered according to Act of Congress, in the year 1872. by 

Harper & Brothers, 

In the Office of the Librarian of Congress, at Washington. 






74 



PREFACE. 



The stereotype plates of my Elements of Analytical Geom- 
etry having become so much worn by long-continued use that 
it was thought desirable to renew them, I have improved the 
opportunity to make a thorough revision of the work. In do- 
ing this, it has been thought best to extend considerably the 
plan of the work, and accordingly I have not merely added a 
third part on Geometry of three dimensions, but have intro- 
duced new matter in nearly every section of the book. I have 
aimed to illustrate every portion of the subject, as far as prac- 
ticable, by numerical examples, generally of the simplest kind, 
the main object being to make sure that the student under- 
stands the meaning of the formulae which he has learned. In 
making this revision I have been favored with the constant 
assistance of Prof. H. A. Newton, who has carefully examined 
every portion of the volume, and to whom I am indebted for 
numerous suggestions both as to the plan and execution of the 
work. It is hoped that the volume in its present form will be 
found adapted to the wants of mathematical students in our 
colleges and higher schools ; and that, if any should desire to 
prosecute this subject further, they will find this volume a good 
introduction to larger and more difficult treatises. 



CONTENTS. 



PART I. 

DETERMINATE GEOMETRY. 

SECTION L 

APPLICATION OF ALGEBRA TO GEOMETRY. 

Papa 

Geometrical Magnitudes represented by Algebraic Symbols 9 

Demonstration of Theorems . 10 

Solution of Problems 12 

SECTION II. 

CONSTRUCTION OF ALGEBRAIC EXPRESSIONS. 

Construction of the Sum or Difference of two Quantities 18 

Product of several Quantities 19 

Fourth Proportional to three Quantities 20 

Mean Proportional between two Quantities 2 L 

Sum or Difference of two Squares 21 

Roots of Equations of the Second Degree 22 

To inscribe a Square in a given Triangle. 26 

To draw a Tangent to two Circles 27 

To divide a straight Line in extreme and mean Ratio 29 



PART II. 

INDETERMINATE GEOMETRY. 
SECTION I. 

CO-ORDINATES OF A POINT. 



Methods of denoting the position of a Point 32 

Abscissa and Ordinate defined — Equations of a Point 33 

Equations of a Point in each of the four Angles 34 

Polar Co-ordinates 35 

Distance of a Point from the Origin 38 

To convert Rectangular Co-ordinates into Polar Co-ordinates 42 

SECTION II. 

THE STRAIGHT LINE. 

Equation of a Straight Line 44 

Four Positions of the proposed Line 46 

Equation of the First Degree containing two Variables 49 

Equation to a Straight Line passing through a given Point 51 

Equation to a Straight Line passing through two given Points 52 

Angle included between two Straight Lines 54 



VI CONTENTS. 

Page 

Condition of Perpendicularity 5G 

Equation to a Straight Line referred to Oblique Axes 59 

Perpendiculars from the Vertices of a Triangle to the opposite Sides 61 

SECTION III. 

TRANSFORMATION OF CO-ORDINATES. 

To change the Origin without altering the Direction of the Axes 64 

To change the Direction of the Axes without changing the Origin 65 

To transform an Equation from Eectangidar to Oblique Co-ordinates 66 

To transform an Equation from Rectangular to Polar Co-ordinates 67 

SECTION IV. 

THE CIRCLE. 

Equation to a Circle when the Origin is at the Centre 69 

Equation to a Circle when the Origin is on the Circumference 71 

Equation to a Circle referred to any Rectangular Axes 72 

Polar Equation to a Circle : 74 

Equation to the Tangent at any Point of a Circle 76 

Equation to the Normal at any Point of a Circle 78 

Co-ordinates of the Points of Intersection of two Circumferences 80 

SECTION V. 

THE PARABOLA. 

Definitions — Curve described mechanically 84 

Equation to the Parabola referred to Rectangular Axes 85 

Equation to the Tangent at any Point of a Parabola 88 

Equation to the Normal at any Point of a Parabola 91* 

Where a Tangent to the Parabola cuts the Axis 92 

Perpendicular from the Focus upon a Tangent 93 

Intersection of a Circle and Parabola 94 

Equation to the Parabola referred to Oblique Axes 95 

Polar Equation to the Parabola 99 

Area of a Segment of a Parabola 100 

SECTION VI. 

THE ELLIPSE. 

Definitions — Curve described mechanically 1 03 

Equation to the Ellipse referred to its Axes 104 

Curve traced by Points 1 06 

Equation when the Origin is at the Vertex of the Major Axis 109 

Squares of two Ordinates as Products of parts of Major Axis 110 

Ordinates of the Circumscribed and. Inscribed Circles Ill 

Equation to the Tangent at any Point of an Ellipse Ill 

To draw a Tangent to an Ellipse through a given Point 113 

Equation to the Normal at any Point of an Ellipse 115 

The Normal bisects the Angle formed by two Radius Vectors 116 

Every Diameter bisected at the Centre 117 

Supplementary Chords parallel to a Tangent and Diameter 120 

Points of Intersection of a Circle and Ellipse 122 

Sum of Squares of two Conjugate Diameters 1 24 

Parallelogram on two Conjugate Diameters 1 25 

Equation to the Ellipse referred to a Pair of Conjugate Diameters 126 



CONTENTS. Vll 

Pnge 

Squares of two Ordinates as Products of parts of a Diameter 127 

Polar Equation to the Ellipse 128 

Directrix of an Ellipse 1 30 

Area of an Ellipse 131 

SECTION VII. 

THE HYPERBOLA. 

Definitions — Curve described mechanically 1 33 

Equation to the Hyperbola 1 34 

Curve traced by Points 137 

Equation when the Origin is at the Vertex of the Transverse Axis 141 

Squares of two Ordinates as Products of parts of the Transverse Axis 141 

Equation to the Tangent at any Point of an Hyperbola 142 

Equation to the Normal at any Point of an Hyperbola 144 

The Tangent bisects the Angle contained by two Radius Vectors 145 

Every Diameter bisected at the Centre 146 

Supplementary Chords parallel to a Tangent and Diameter 148 

Properties of Conjugate Diameters 150 

Difference of Squares of Conjugate Diameters 152 

Equation to the Hyperbola referred to two Conjugate Diameters 1 53 

Squares of two Ordinates as the Rectangles of the Segments of a Diameter. . 155 

Polar Equation to the Hyperbola 155 

Directrix of an Hyperbola 158 

Asymptotes of the Hyperbola 160 

Tangents through the Vertices of two Conjugate Diameters 162 

Equation to the Hyperbola referred to its Asymptotes 165 

Equation to the Conjugate Hyperbola 166 

Equation to the Tangent referred to Asymptotes 168 

Intersection of Tangent with the Axes 169 

SECTION VIII. 

GENERAL EQUATION OF THE SECOND DEGREE. 

The Term containing the first Power of the Variables removed 170 

The Term containing the Product of the Variables removed 171 

Discussion of the resulting Equation 1 72 

Lines represented by the general Equation of the second Degree 175 

Equation to the Conic Sections referred to same Axes and Origin 178 

SECTION IX. . 

LINES OF THE THIRD AND HIGHER ORDERS. 

General Equation of the Third Degree 180 

Equations of the Eourth Degree 182 

SECTION X. 

TRANSCENDENTAL CURVES. 

Cycloid defined 181 

Equation of the Cycloid 1 85 

Logarithmic Curve defined 186 

Curve of Sines, Tangents, etc 187 

Spirals — Spiral of Archimedes — its Equation 188 

Hyperbolic Spiral — its Equation 191 

Logarithmic Spiral — its Equation 102 



Vlll CONTENTS. 



PART III. 

GEOMETRY OF THREE DIMENSIONS. 

SECTION I. 

OF POINTS IN SPACE. 

Page 

Position of a Point in Space denoted 194 

Distance of any Point from the Origin .-. . 197 

SECTION II. 

THE STRAIGHT LINE IN SPACE. 

Equation to a Straight Line in Space 199 

Equation to a Straight Line passing through a given Point 201 

Equation to a Straight Line parallel to a given Line 204 

SECTION III. 

OF THE PLANE IN SPACE. 

Equation to a Plane 206 

Equation of the Plane which passes through three given Points 208 

Conditions which must subsist in order that two Planes may be parallel 210 

Equation of a Plane perpendicular to a given Straight Line 211 

SECTION IV. 

OF SURFACES OF REVOLUTION. 

Solid of Revolution defined 21-1 

Equation to the Surface of a Right Cylinder 214 

Equation to the Surface of a Right Cone 215 

Equation to the Surface of a Prolate Spheroid 216 

Equation to the Surface of an Oblate Spheroid 217 

Equation to the Surface of an Hyperboloid 218 

Curve which results from Intersection of a Cylinder and Plane 220 

Curve which results from Intersection of a Cone and Plane 221 

Curve which results from Intersection of a Spheroid and Plane 227 

Curve of Intersection of a Plane and Paraboloid 228 

Summary of preceding Results 230 

SECTION V. 
GENERAL EQUATION OF THE SECOND DEGREE BETWEEN THREE VARIABLES. 

Classification of Surfaces represented by the general Equation 233 

Particular Cases of the general Equation 235 

Section of a Surface of the second Degree by a Plane 236 



APPENDIX. 

On the graphical Representation op natural Laws 239 



ANALYTICAL GEOMETRY. 



PART I. 
DETERMINATE GEOMETRY. 

SECTION I. 
APPLICATION OF ALGEBRxV TO GEOMETRY, 

1. We have seen in Geometry (pages 40, 69, and 162) that 
all geometrical magnitudes,- including angles, lines, surfaces, 
and solids, may be expressed either exactly or approximately 
by numbers, and for this purpose it is only necessary to take 
one of these magnitudes as the unit of measure. If we denote 
by «, J, and c the number of linear units contained in the ad- 
jacent edges of a rectangular parallelopiped, then will db, ac, 
be denote the magnitude of three of its faces, and abc will de- 
note its volume. 

2. In like manner, every geometrical magnitude may be rep- 
resented by algebraic symbols, and the relations between dif- 
ferent magnitudes, or different parts of the same figure, may 
also be denoted by symbols. We may then operate upon these 
representatives by the known methods of Algebra, and thus 
deduce relations before unknown ; and since the operations 
are generally very much abridged by the use of algebraic sym- 
bols, the algebraic method has many advantages over the geo- 
metrical. This method is applicable either to the solution of 
problems or to the demonstration of theorems. 

3. Geometrical problems may be divided into two classes : 
determinate and indeterminate. Determinate problems are 
those in which the number of independent equations is equal 
to the number of unknown quantities, and therefore the un- 

A2 



10 ANALYTICAL GEOMETRY. 

known quantities can have but a finite number of values. In- 
determinate problems are those in which the number of inde- 
pendent equations is less than the number of unknown quan- 
tities involved, and therefore the unknown quantities may have 
an infinite number of values. 

4. If it is required to determine the magnitude of certain 
lines from the knowledge of several other lines connected with 
the former in the same figure, w T e first draw a figure which rep- 
resents all the parts of the problem, both those which are given 
and those which are required to be found. We denote both 
the known and unknown parts of the figure, or as many of 
them as may be necessary, by convenient symbols. We then 
observe the relations which the several parts of the figure bear 
to each other, from which, by the aid of the proper theorems 
in Geometry, we derive as many independent equations as 
there are unknown quantities employed. By solving these 
equations we obtain expressions for the unknown quantities in 
terms of the known quantities. 

If a theorem is to be demonstrated, we express by algebraic 
equations the relations which exist between the different parts 
of the figure, and then transform these equations in such a 
manner as to deduce an equation which expresses the theorem 
sought. 



■~o 



5. In order to illustrate these principles, let it be required to 
deduce the various properties of a right-angled triangle from 
the principles that two equiangular triangles have their ho- 
mologous sides proportional, and that the perpendicular drawn 
from the right angle of a right-angled triangle to the hypothe- 
nuse divides the whole triangle into similar triangles. 

Let the triangle ABC be right angled at A ; 
from A draw AD perpendicular to BC, and let 
us put BC = a, AC — Z>, AB = c, AD = A, BD = m 9 
c and DC — ?i. Then, by similar triangles, we 
have the proportions 




APPLICATION OF ALGEBRA TO GEOMETRY. 11 



a:b::b:n ) ( b 2 = a?i. (1) 

a: c:\c\m > whence we deduce ■< & — am. (2) 



•j c 2 = < 
a: c:\b\h ) I be = ah. (3) 

Moreover, we have a = m + n. (4) 

These four equations involve the various properties of right- 
angled triangles, and these properties may be deduced by suit- 
able transformations of these equations. * 

1st. Equations (1) and (2), or rather the proportions from 
which they are deduced, show that each side about the right 
angle is a mean proportional between the entire hypothennse 
and its adjacent segment. 

2d. By adding equations (1) and (2) member to member, we 
obtain 

b 2 -f c 2 — am + an — a(m + n) ; 
whence, from equation (4), we obtain b 2 + c 2 =a 2 ; that is, the 
square of the hypothennse is equal to the sum of the squares 
of the other two sides of the triangle, 

3d. By multiplying equations (1) and (2) member by mem- 
ber, we obtain 

b 2 c 2 — a 2 mn. 
But from equation (3) we have also b 2 c 2 — a 2 h 2 . 
Hence a 2 ?n?i = a 2 h 2 y or h 2 ~mn; that is, 

m:h::h: n, 
or, the perpendicular drawn from the vertex of the right an- 
gle upon the hypothennse is a mean proportional between the 
two segments of the hypothennse. 

4th. By dividing equation (1) by equation (2) member by 

member, we obtain 

b 2 an 72 9 

■—= — / or o 2 :c 2 ::n: m : 

c 2 am 

that is, the squares described upon the sides about the right 

angle are proportional to the segments of the hypothennse. 

Thus we see that every equation deduced from the equations 

(1), (2), (3), and (4), when translated into geometrical language, 

is a geometrical theorem. 



12 ANALYTICAL GEOMETRY. 

6. The four equations of the preceding article contain six 
quantities, of which, when a certain number are given, it may 
be required to deduce the values of the other quantities. 

Suppose we have given the hypothenuse BC, and the per- 
pendicular AD, and it is required to determine the other two 
sides of the triangle, as also the two segments of the hypothe- 
nuse. 

We have already found b 2 + c 2 = a 2 , 
and from equation (3) w r e have 2bc—2ah. 
By adding and subtracting successively, we obtain 

(b + c) 2 = a 2 + 2ah; 
and (b — c) 2 — a 2 — 2ah. 

whence b+c= \/a 2 + 2ah ; b — c~ yV — 2ah. 

Knowing the sum and difference of the two sides b and <?, by a 
well-known principle (Alg., p. 89) we obtain 
the greater side b = ^^a 2 -\-2ah + ^^a 2 —2ah, 
the less side c — J i/a 2 + 2ah — \ \/a 2 — 2ah. 

Since a, b, and c are now known quantities, the two segments 
are given by equations (1) and (2). 

The preceding principles will be further illustrated by the 
following examples : 

Ex. 1. The base and sum of the hypothenuse and perpendic- 
ular of a right-angled triangle are given, to find the perpendic- 
%dai\ 

Let ABC be the proposed triangle, right angled 
at B. Represent the base AB by £, the perpendic- 
ular BC by #, and the sum of the hypothenuse and 
perpendicular by s; then the hypothenuse will be 
represented by s-x. 

By Geom.,B.IV.,Pr.ll, AB^ + BC^AC 2 ; 
l2 + x 2 = (s—x) 2 = s 2 —2sx + x 2 . 
b 2 =s 2 —2sx, 




2s ' 
that is, in any right-angled triangle, the perpendicular is equal 
to the square of the sum of the hypothenuse and perpendicu- 



APPLICATION OF ALGEBRA TO GEOMETRY. 



13 




lar, diminished by the square of the base, and divided by twice 
the sum of the hypothenuse and perpendicular. 

Thus, if the base is 3 feet, and the sum of the hypothe- 

S 2 — b 2 

nuse and perpendicular 9 feet, the expression — — becomes 

9 2 — 3 2 . 

- — q-=4, the perpendicular. 

Z x J 

Ex. 2. The base and altitude of any triangle are given, and 
it is required to find the side of the inscribed square. 

Let ABC represent the given triangle, 
and suppose the inscribed square DEFG 
to bt3 drawn. Represent the base AB by 
b, the perpendicular CH by h, and the side 
of the inscribed square by x ; then will 
CI be represented by h—x. 

Then, because GF is parallel to the base AB, we have by 
similar triangles (Geom., B. IV., Pr. 16), 

AB:GF::CII:CI; 
that is, b:x::h: h—x, 

whence bh —bx — hx ; 

bh 

0T > X ~~J+h ; 

that is, the side of the inscribed square is equal to the product 

of the base and height divided by their sum. 

Thus, if the base of the triangle is 12 feet, and the altitude 
6 feet, the side of the inscribed square is found to be 4 feet. 

Ex. 3. The base and altitude of any triangle are given, and 
it is required to inscribe within it a rectangle whose sides shall 
have to each other a given ratio. 

Let ABC be the given triangle, and sup- 
pose the required rectangle to be inscribed 
within it. Represent the base AB by b, 
the altitude CH by h, the altitude of the 
rectangle DG by x, and its base DE by y ; 
also let x : y : : 1 : n ; or y — nx. 

Then, because the triangle CGF is similar to the triangle 
CAB, we have 




14: ANALYTICAL GEOMETRY. 

AB:GF::CII:CI; 
that is, b : y : : h : h—x ; 

whence bh — bx—hy. 

But, since y — nx y we have 

bh—bx — hnx; 

whence a 



If we suppose ?i equals unity, in which case the rectangle 
becomes a square, the preceding result becomes identical with 
that in Example 2. 

Ex. 4. It is required to divide a straight line in extreme 
and mean ratio ; that is, into two parts such that one of them 
shall be a mean proportional between the ivhole line and the 
other part. 

Suppose the .problem to be solved, and that C is such a point 

of the line AB that w r e have the proportion , 

AB:AC::AC:CB. A c B 

Put AB=a, AC=#, whence CB = a— x. 

The preceding proportion will then become 
a.x».x.a —— x j 
whence x 2 =a 2 —ax, 

which equation, being solved, gives 

Of these two values obtained by the solution of the equation, 
the first is the only one which satisfies the enunciation of the 
problem ; for the second is numerically greater than a, and 
therefore can not represent a part of the given line. We shall 
consider hereafter the geometrical signification of this equa- 
tion. 

Ex. 5. It is required to determine the side of an equilateral 
triangle described about a circle whose diameter is given. 

Suppose ABC to be the required triangle described about a 
circle whose diameter is given. Draw AE perpendicular to 
BC, and join DC. Represent FE by d, and CE by x. The two 
triangles ACE, CDE are similar, for each contains a right an- 




APPLICATION OF ALGEBKA TO GEOMETRY. 15 

gle, and the angle CAE is equal to the angle 
DCE. Hence we have the proportion 

AC:EC::DC:DE. 
But AC is double of EC; therefore DC is 
double of DE, or is equal to d. 
Now DC 2 -DE 2 =EC 2 , 

d 2 

or d 2 —-r=x 2 , 

4 

whence x=^d\/3, 

or 2x—d\/3; 

that is, the side of the circumscribed triangle is equal to the 

diameter of the circle multiplied by the square root of 3. 

Ex. 6. Given the base h and the difference d between the hy- 

pothenuse and perpendicular of a right-angled triangle, to find 

the perpendicular. 

A b 2 -d 2 
Am. — j-, 

Ex. 7. Given the hypothenuse A of a right-angled triangle, 
and the ratio of the base to the perpendicular, as m to n, to 
find the perpendicular. 

Ans. 



-yjiin? + n 2 

Ex. 8. Given the diagonal d of a rectangle, and the perime- 
ter 4j9, to find the lengths of the sides. 

/d 2 

Ex. 9. If the diagonal of a rectangle be 10 feet, and its pe- 
rimeter 28 feet, what are the lengths of the sides ? 

Am. 

Ex. 10. From any point within an equilateral triangle, per- 
pendiculars are drawn to the three sides. It is required to find 
the sum, s, of these perpendiculars. 

# Arts, s — altitude of the triangle. 

Ex. 11. Given the lengths of three perpendiculars, #, b, and 



16 ANALYTICAL GEOMETRY. 

c, drawn from a certain point in an equilateral triangle to the 
three sides, to find the length of the three sides. 

yo 
Ex. 12. Given the difference, d, between the diagonal of a 
square and one of its sides, to find the length of the sides. 

Ans. d+dV2. 
Ex. 13. In a right-angled triangle, the lines a and b, drawn 
from the acute angles to the middle of the opposite sides, are 
given, to find the lengths of the sides. 



Ans. ^\J^-tA, and 2y ■ 



'4a 2 -6 a 



15 ' V 15 ' 

Ex. 14. In a right-angled triangle, having given the hypothe- 
nuse (a), and the difference between the base and perpendicu- 
lar (2d), to determine the two sides. 



/a* -2d 2 , 7 , la 2 -2d 2 7 
Ans. y — - \-d, and y — d, 

Ex. 15. Having given the area (c) of a rectangle inscribed 
in a triangle whose base is (b) and altitude (a), to determine 
the height of the rectangle. 



. a , la 2 c 



ae 
2~ V 4 J' 

Ex. 1G. Having given the ratio of the two sides of a triangle, 
as m to n, together with the segments of the base, a and b, 
made by a perpendicular from the vertical angle, to determine 
the sides of the triangle. 



/a 2 -b 2 . ft 
Ans. m\ — 5, and n\ - 



-n* v m' — n' 

Ex. 17. Having given the base of a triangle (2a), the sum of 
the other two sides (2s), and the line (c) drawn from the verti- 
cal angle to the middle of the base, to find the sides of the tri- 
angle. 



-&* 



Ans. s± i/a 2 + c 2 —s 2 . 
Ex. 18. Having given the two sides (a) and (b) about the ver- 
tical angle of a triangle, together with the line (c) bisecting 



APPLICATION OF ALGEBJRA TO GEOMETEY. 17 

that angle and terminating in the base, to find the segments of 
the base. 



. lab—<? t 7 la 

Ans. ay — -7 — , and o\J - 



ab V ab 

Ex. 19. The sum of the two legs of a right-angled triangle is 
s, and the perpendicular let fall from the right angle upon the 
hypothenuse is a. What is the hypothenuse of the triangle ? 

A?is. -\/s 2 +d 2 —a. 
Ex. 20. Determine the radii of three equal circles, described 
in a given circle, which touch each other, and also the circum- 
ference of the given circle whose radius is E. 

Ans. TL{2V3-3). 



18 ANALYTICAL GEOMETRY. 



SECTION II. 

CONSTRUCTION OF ALGEBRAIC EXPRESSIONS. 

7. The construction of an algebraic expression consists in 
finding a geometrical figure which may be considered as rep- 
resenting that expression ; that is, a figure in which the parts 
shall have the same geometrical relation as that indicated in 
the algebraic expression. 

The elementary expressions, to which all algebraic expres- 
sions not exceeding the second degree may be reduced, are six 
in number, viz., 

, , , ab a? 

x=a — o + c—a.etc, x= — , x=— 9 

c 7 <r 

x—Vab y x=Va z + b 2 , x=Va 2 —b 2 ; 

where a, b, c, etc., express the number of linear units contained 
in the given lines. 

Problem I. To construct the expression x — a + b. 

The symbols a and b, being supposed to stand for numerical 

quantities, may be represented by lines. The length of a line 

is determined by comparing it with some known standard, as 

an inch or a foot. If the line AB contains the standard unit 

, l_ a times, then AB may be taken to repre- 

c sent a. So, also, if BC contains the stand- 
ard unit b times, then BC may be taken to represent b. There- 
fore, in order to construct the expression a + b, draw an indefi- 
nite line AD. From the point A lay off a distance AB equal 
to a, and from B lay off a distance BC equal to b ; then AC will 
be a right line representing a + b. 

Problem II. To construct the expression x — a — b. 

, Draw the indefinite line AD. From 

^ B D t j ie p i n t A lay off a distance AB equal to 

a, and from B la}' off a distance BC, in the direction toward A, 



CONSTRUCTION OF ALGEBRAIC EXPRESSIONS. 19 

equal to b ; then will AC be the difference between AB and 
BC ; consequently, it may be taken to represent the expression 
a— b. 

Problem III. To construct the expression 
x=a—b + c—d+e. 

This expression may be written 

x—a+c+e— (b+d). 

To obtain an expression for a+c+e, draw an indefinite line 
AX, and from A set off AB = &, . , , , . . 

f rom B set off BC = c, from C set * * B E c D x 
off CD = e; then &D = a+c+e. 

Then set off from D toward A, DE = b ; from E set off 
EF=d; then T>F = b + d. 

Hence AF=a+c+e—(b+d)—x. 

In a similar manner we may construct any algebraic express 
sion consisting of a series of letters connected together by the 
signs + and — . 

In like manner we may construct the expressions x=3a, 

K7 in • a 2d 3a / 

x — ido^ etc.; also the expressions x=-, x= — , x=—, etc. 

Problem IY. To construct the expression x=ab. 

Let ABDC be a rectangle of which the side AB contains 
the standard unit a times, and the side AC contains the same 
unit b times. If through the points E, 
F, etc., we draw lines parallel to AC, i — i — \ — i — \ — i 

and through the points G, H, etc., we h X 

draw lines parallel to AB, the rectangle G 1 

will be divided into square units. In A — -L— ^ — I — I — I 
the first row, AGIB, there are a square 
units ; in the second row, GTIKI, there are also a square units ; 
and there are as many rows as there are units in AC. There- 
fore the rectangle ABDC contains a x b square units, or the 
rectangle may be considered as representing the expression ab. 

An algebraic expression of two dimensions may therefore 
be represented by a surface. 



20 ANALYTICAL GEOMETKY. 

Problem V. To construct the expression x—abc. 

Let there be a rectangular parallelopiped whose three adja- 
cent edges contain the standard unit respectively a, 5, and c 
times ; then, dividing the solid by planes parallel to its sides, 
we may prove that the number of solid units in the figure is 
axbxc, and consequently the parallelopiped may be consider- 
ed as representing the expression abc. 

An algebraic expression of three dimensions may therefore 
be represented by a solid. 

Problem VI. To construct the expression x— — . 

From this equation we derive the proportion 

that is, x is a fourth proportional to the three given lines c y a, 

and b. 

To obtain an expression for x, draw two 
lines, AB, AC, making any angle with 
each other. From A, upon the line AB, 
lay off a distance AD=c, and AB=a, 
and upon the line AC lay off a distance 

AE=b. Join DE, and through B draw BO parallel to DE; 

then will AC be equal to x. 

For, by similar triangles, we have 

AD:AB::AE:AC, 

or c : a : : b : AC. 

Hence AC=: — =x. 

c 

ct ci ^ h 

The expression x=— , or x— , may be constructed in the 

c o 

same manner, since x is a fourth proportional to the three lines 

c, a, and a. 

Problem VII. To construct the expression x=—. 

This expression can be put under the form 

__ab c 

X — 7" X — • 

a e 




CONSTRUCTION OF ALGEBRAIC EXPRESSIONS. 21 

First find a fourth, proportional m to the three quantities 

7 

d, a, and b, as in Prob. VI. This gives us m = -j. The pro- 

mC 

posed expression then becomes — , which may be constructed 

in a similar manner. 

In like manner more complicated expressions may be con- 

a 3 b 2 c 
structed ; as 1o ~ 9 . 

Problem VIII. To construct the expression x— Vab. 

The expression Vab denotes a mean proportional between 
a and b ; for we have 

x 2 = axb; or a:x::x:b. 

To construct this expression, draw an in- j) 

definite straight line, and upon it set off 
AB = a, and BC = b. On AC as a diameter, / 
describe a semicircumference, and from B [_ 
draw BD perpendicular to AC, meeting the A B 

circumference in D ; then BD is a mean proportional between 
AB and BC (Geom., Bk. IV., Prok23, Cor.). Hence BD is a 
line representing the expression Vab—x. 

Problem IX. To construct the expression x— Va 2 +b 2 . 

This expression represents the hypothenuse of a right-angled 
triangle, of which a and b are the two sides c 

about the right angle. 

Draw the line AB, and make it equal to a ; 
from B draw BC perpendicular to AB, and 
make it equal to b. Join AC, and it will rep- 
resent the value of Va 2 + b 2 ; since AC 2 =AB 2 +BC 2 (Geom., 
Bk. IV., Prob. 11). 

Problem X. To construct the expression x—Vd l — b 2 . 

This expression represents one of the sides of a right-angled 
triangle, of which a represents the hypothenuse, and b the re- 
maining side. 

Draw an indefinite line AB ; at B draw BC perpendicular 
to AB, and make it equal to b. With C as a centre, and a 




22 ANALYTICAL GEOMETRY. 

C radius equal to #, describe an arc of a circle 
cutting AB in D ; then will BD represent 
the expression Va 2 —b 2 . For 

BD 2 =DC 2 --B C 2 =a 2 -6 2 . 
D\ B Whence RT) = Vd z -b 2 =x. 




Problem XL To construct the expression x—Va 2 -\-b 2 —c 2 . 

Put a 2 + b 2 =d 2 , and construct d as in Prob. IX.; then we 
shall have x = Vd 2 — c 2 , 

which may be constructed as in Prob. X. 

In the same manner we may construct the expression 
x—Vd z — b 2 + c 2 — d 2 +e 2 — , etc. 

By methods similar to the preceding the following expres- 
sions may be constructed : 

1. x—Va 2 + ab. 4. x=Vd 2 — bc. 

2. x=Vab + cd. 5. x — a 2 -\-ab. 

labc a 3 

3. x = \J—. 0. *=-. 

Problem XII. To construct the roots of the four forms of 
equations of the second degree (Alg., Art. 277). 

In the equation x 2 ±px— ±:q, 

x 2 and jpx represent surfaces (Prob. IY.) ; q must therefore rep- 
resent a surface. We will suppose this surface transformed 
into a square (7c 2 ), and, to avoid misapprehension, will write the 
general equation of the second degree 

X 2 ±pX=dtk 2 . 

First form. The first form x 2 +px=7c 2 gives for x the two 

p /p 2 p lp 2 

values x = —- } + \J ~r + ^ 2 an ^ x = -^-y^ + 7c 2 . 

Draw the line AB, and make it equal 
to 7c. From B draw BC perpendicular to 

p 

AB, and make it equal to ^r. Join A and 

C ; then, as in Prob. IX., AC will repre- 
sent the value of \ ~t + 7c 2 . 




CONSTRUCTION OF ALGEBRAIC EXPRESSIONS. 23 

With C as a centre, and CB as a radius, describe a circle 

cutting AC in D, and AC produced in E. For the first value 

of x the radical is positive, and is set off from A toward C ; 

V 
then —-: is set off from C to D, and AD, which equals 



4 



"\e-l 



represents the first value of x, measured from A to D. 

For the second value of x we begin at E, and set off EC 

equal to — £ ; we then set off the minus radical from C to A ; 

then EA, measured from E to A, represents the second value 
of x. 

Second form. The second form x 2 —jjx = k 2 gives for x the 
two values 

*=|+V^+£ 2 aiid*=|-.V?+*». 

Fp 2 

Construct as before AC = \ -^r+Jc 2 \ then from C lay off CE 
equal to"^, and the first value of x will be represented by AE, 

measured from A to E. 

p 

From D lay off DC equal to%; then from C in a contrary 



lp 2 
direction lay off CA equal to \J x+& 2 , and the second value 

of x will be represented by DA, measured from D to A. 

Third form. The third form x 2 +px=—k 2 gives for x the 
two values 



Draw an indefinite line FA, and from 

p 

any point, as A, set off a distance AB= — — . 

p V 

We set off this line to the left, because % is * D ^ <ea 




2i ANALYTICAL GEOMETRY. 

negative. At B draw EC perpendicular to FA, and make it 
equal to k. From C as a centre, with a radius equal to - ^, de- 
scribe an arc of a circle cutting the line FA in D and E. Join 

/pi 
CD, and we shall have ED or EE equal to yx— ^ 2 - 

The first value of x will be represented by — - AB+BE, which 
is equal to — AE. The second value of x will be represented 
by —AB—BD, which is equal to —AD; so that both of the 
roots are negative, and are measured from A toward the left. 

Fourth form. The fourth form # 2 — px=*z— h 2 gives for x 
the two values 



4 + Vf-^ a ? a*=f-vf 



a. :ir 



Set off AB equal to ~ from A toward the 
right. We set it off toward the right be- 
^ cause g is positive. Then construct the rad- 



ical part of the value of x as for the third form. To AB we 
acid BD, which gives AD for the first value of x; and from 
AB we subtract BE, which gives AE for the second value of x. 
Both values are positive, and are measured from A toward the 
right. 

Equal roots. If the radius CE be taken equal to CB, that 
p 
is, if k is equal to tt, the arc described with the centre C will 

not cut the line AF, but will touch it at the point B, the two 
points D and E will unite, the radical part of the value of x be- 
comes zero, and the two values of x become equal to each other. 

Imaginary roots. If the radius of the circle described with 
c the centre C be taken less than CB, it will not 

meet the line AF. In this case I 2 is numerically 



j 



X- — B greater than |- 3 and the radical part of the value 



A 

of x becomes imaginary 



CONSTRUCTION OF ALGEBKAIC ^EXPRESSIONS. 25 

8. Every algebraic expression admitting of geometrical con- 
struction must have all its terms homogeneous (Alg., Art. 33) ; 
that is, each term must be of the same degree. The degree of 
any monomial expression is the number of its literal factors. 
If, however, the expression have a literal divisor, its degree is 
the number of literal factors in the numerator diminished by 
the number in the denominator. Thus the expressions x, 

ab abc „ . n t , . . n a 2 b abed 

— , -j- are 01 the nrst decree ; the expressions x ? \ — , — -r- are 
e ■* de ° ' x ' c ' ejt 

of the second degree. In order that an algebraic expression 

may admit of geometrical construction, each term must either 

be of the first degree, and so represent a line ; or, secondly, 

each must be of the second degree, and so represent a surface ; 

or, thirdly, each must be of the third degree, and denote a solid, 

since dissimilar geometrical magnitudes can neither be added 

together nor subtracted from each other. 

It may, however, happen that an expression really admitting 
of geometrical construction appears to be not homogeneous ; 
but this result arises from the circumstance that the geomet- 
rical unit of length, having been represented in the calculation 
by the numeral unit 1, disappears from all algebraic expres- 
sions in which it is either a factor or a divisor. To render 
these results homogeneous, it is only necessary to restore this 
factor or divisor which represents unity. 

Thus, suppose we have an equation of the form 

• x = ab + c. 

If we put I to represent the unit of measure for lines, we may 
change it into the homogeneous equation 

lx — ab-\-cl^ 

ab 
or x = -j-+c, 

which is easily constructed geometrically. 

Suppose the expression to be constructed to be of the form 

x - b-2c+3' 
B 



26 A^LYTICAL GEOMETKY. 

Since one of tlie terms of the numerator is of the second de- 
gree, each of the other terms of the numerator should be made 
of the same degree, and each term in the denominator should 
be made of the first degree ; so that, introducing the linear 
unit Z, the expression to be constructed is 

a 2 + 3lb-2l 2 
x - b-2c+3l * 
The denominator of this fraction may be constructed by Prob. 
III. If we represent the denominator by m, the expression 
may be written 

~~m m m* 
each of which terms may be constructed by Prob. YI. 

The following examples will show how an algebraic solution 
of a problem may be converted into a geometrical solution. 

Problem XIII. Having given the base and altitude of any 
triangle, it is required to find the side of the inscribed square 
by a geometrical construction. 

AVe have found, on page 13, the side of the inscribed square 

to be equal to 7,7 / that is, it is a fourth proportional to 

b + h, b and h. 

In order to construct this expression, produce the base AB 
until BL is equal to the altitude h; through L draw LM par- 
allel to BC, meeting CM drawn through 
C parallel to AB. Join AM, and let it 
meet BC in F; draw FE perpendicular 
to AB, and it will be the required line. 
~S x L Draw MN perpendicular to AL. 
By similar triangles we have 

AL:AB::LM:BF::MN:FE; 
that is, b + h;b::h:FE; 

whence YE= 1 — 7 =x: 

o + h ' 

and therefore EF is equal to a side of the inscribed square. 

Example 3, page 13, may be constructed in a similar manner 

by laying off BL equal to nh. 




CONSTRUCTION OF ALGEBRAIC EXPRESSIONS. 



27 



Problem XIV. It is required to draw a straight line tan- 
gent to two given circles situated in the same plane. 

Since the two circles are given both in extent and posi- 
tion, we know their radii and the distance between their cen- 
tres. 

Let C, C be the centres of the two circles, CM, CM' their 
radii. Denote the radius CM of the first circle by r, that of 
the second CM 7 by r\ and the distance between their centres 
CC by a. Suppose that MM' is the required tangent ; pro- 
duce this line to meet CC produced in T, and denote the dis- 
tance CT by x. 

There are two cases : 

Case First When the tangent does not pass between (he 
circles. 

Draw the radii CM, CM' to 
the points of tangency; the an- 
gles CMT, CM'T will be right 
angles, and the triangles CMT, 
C'M'T will be similar. Hence 
we shall have the proportion 




CM:C'M'::CT:C'T, 

r : r' \ : x : x—ei; 
rx—ra—r'x, 
ar 



x-- 



w> 



or 
whence 

and 

r—r 

from which we see that CT or x is a fourth proportional to 
r—r', a and r. 

To obtain x by a geometrical construction, through the cen- 
tres C and C draw any two 
parallel radii CNjC'N', on the 
same side of CC. Through 
N and W draw the line NK 7 , 
and produce it to meet CC 
produced in T. CT will be 
the line represented by x. 

For through N' draw WB parallel to CT; then ND will 







28 



ANALYTICAL GE0METKY. 



represent r—r\ and NT) will be equal to a ; and by similar 

DN:DN'::CN:CT, 

r-r':a::r:GT; 



triangles we have 
or 



whence 



CT: 



ar 



/p ___ /p' 



-X. 



Therefore a line drawn from T, tangent to one of the cir- 
cles, will also be tangent to the other ; and, since two tangent 
lines can be drawn from the point T, we see that this first case 
of the proposed problem admits of two solutions. 

Cor. If we suppose the radius r of the first circle to remain 
constant, and the smaller radius r' to increase, the difference 
r— r r will diminish ; and, since the numerator ar remains con- 
stant, the value of x will increase ; which shows that the nearer 
the two circles approach to equality, the more distant is the 
point of intersection of the tangent line with the line joining 
the centres. When the two radii r and r r become equal, the 
denominator becomes zero, the value of x becomes infinite, and 
the two tangents are parallel. 

If we suppose r' to increase so as to become greater than r, 
the value of x becomes negative, which shows that the point T 
is on the left of the two circles. 

Case Second. When the tangent passes between the circles. 

In this case, as in the other, the 
lines CM and CM' are parallel; 
hence the triangles CMT, C'M'T 
are similar, and we have the pro- 
portion 
CMiC^M'riCTiC'T, 
r : r r \ : x : a—x: 




or 



whence 



ar 



X — 




~ r+r'' 

To construct this expression, through 
the centres C and C' draw any two 
parallel radii GN", C'N', lying on dif- 
ferent sides of CC; join the points 
NN', sjid through T, where this line 



CONSTRUCTION OF ALGEBRAIC EXPRESSIONS. 29 

intersects CC, draw a line tangent to one of the circles. It 
will be a tangent to the other. 

For through W draw NT) parallel to CC 7 , and meeting CN 
produced in D. From the similar triangles NCT, NDN' we 
have the proportion 

ND:DN'::NC:CT, 
or r+ r':a::r :CT; 

CLV 

whence CT = — — y—x. 

r+r 

Cor. The value of x is positive for all values of r and iff 
when r=r\ the value of x reduces to ~. 

If each circle is wholly exterior to the other, there may 
therefore be two exterior tangents and two interior tangents, 
in which case the problem admits oifoar solutions. 

If the two circles touch each other externally, the two inte- 
rior tangents unite in one, and the problem admits but three 
solutions. 

If the two circles cut each other, the interior tangents are 
impossible, and the problem admits but two solutions. 

If the two circles touch each other internally, the two exte- 
rior tangents unite in one, and the problem admits but one 
solution. 

If one circle is wholly interior to the other, no tangent line 
can be drawn, and no solution of the problem is possible. 

The general values of x already found undergo changes cor- 
responding to the changes here supposed in the position of the 
two circles. 

Problem XV. To divide a straight line in extreme and 
mean ratio. 

"We have found, in Example 4, page 14, 

To construct the first value of % y make AB = a; at B erect 
the perpendicular BC=^, and join AC. 



30 ANALYTICAL GEOMETRY. 

Then, as inProb. 9, page 21, 

From C as a centre, with a 
radius CB = ^, describe a cir- 
cumference cutting AC in D and AC produced in E. From 
AC take CD^, and we have 




AD=AC-CD = \A 2 +| 2 -! 

To construct the second value of x. 

ct 
From E set off EC towards tlie left equal to „, and from 



V« 2 +x 



also towards the left set off CA equal to \J a 2 + -r. Then EA, 
measured from E to A, will represent 



'=-2-V« 2 +4- 



With A as a centre, and AD as a radius, describe the arc 
DF. The line AB will be divided in the required ratio at F, 
and AF will be the greater part. 

The second value of x~ — AE is numerically greater than 
AB. It can not, then, form a part of AB, and is not an an- 
swer to the question in the form here proposed. 

Each value of x may, however, be regarded as the solution 
of the more general problem, " Two points A and B being 
given, to find, on the indefinite line that passes through them, 
a third point F, such that the distance AF shall be a mean pro- 
portional between the distances AB and BF." To this problem 
there are evidently two solutions, F on the right of A being 
one of the points, and F' on the left of A is the other. 

9. From the preceding examples we perceive that the solu- 
tion of a geometrical problem by the aid of Algebra consists 
of three principal parts : 



CONSTRUCTION OF ALGEBRAIC EXPRESSIONS. 31 

1st. To translate the problem into algebraic language, or to 
reduce it to an equation. 

2d. To solve the equation or equations. 

3c/. To construct geometrically the algebraic expressions ob- 
tained. 

Frequently it becomes necessary to add a fourth part, whose 
object is the discussion of the problem, or an examination of 
all the circumstances relating to it. 



32 ANALYTICAL GEOMETKY. 



PART II 

INDETERMINATE GEOMETRY. 






SECTION I. 
CO-OEDINATES OF A POINT. 

10. The object of the second branch of Analytical Geometry 
is to determine the algebraic equations by which known lines 
and curves may be represented, and from these equations to 
deduce their geometrical properties ; and conversely, having 
given the equations, to determine the lines and curves which 
they represent. 

11. To determine the position of a point in a plane. The 
position of a point in a plane may be denoted by means of its 
distances from two given lines which intersect one another. 

Thus, let AX, AY be two assumed straight 
lines which intersect in any angle at A, and 
let P be any point in the same plane ; then, 
if we draw PB parallel to AY, and PC par- 
allel to AX, the position of the point P will 
-x be determined by means of the distances 
PB and PC. 
The two lines AX, AY, to which the position of the point P 
is referred, are called axes, and their point of intersection, A, is 
called the origin. The distance AB, or its equal CP, is called 
the abscissa of the point P ; and BP, or its equal AC, is called 
the ordinate of the same point. Hence the axis AX is called 
the axis of abscissas, and AY is called the axis of ordinates. 
The abscissa and ordinate of a point, when spoken of togeth- 
er, are called the co-ordinates of the point, and the two axes 
are called axes of co-ordinates, or co-ordinate axes. 

A system of axes may be either rectangidar or oblique ; 
that is, the angle YAX may be either a right angle or an 



/ 



CO-ORDINATES OF A POINT. 33 

oblique angle. Rectangular axes are ordinarily most conven- 
ient, and will generally be employed in this treatise. 

An abscissa is usually denoted by the letter x, and an ordi- 
nate by the letter y / and hence the axis of abscissas is often 
called the axis of x, and the axis of ordinates the axis of y. 

The abscissa of any point is its distance from the axis of 
ordinates measured on a line parallel to the axis of abscissas. 

The ordinate of any point is its distance from the axis of 
abscissas measured on a line parallel to the axis of ordinates. 

12. Equations of a point. The position of a point may be 
determined when its co-ordinates are known. For, suppose the 
abscissa of the point P is equal to 5, and its ordinate is equal 
to 4. Then, to determine the position of the 
point P, from the origin A lay off on the axis 
of abscissas a distance AB equal to 5 units 
of length, and through B draw a line paral- 
lel to the axis of ordinates. On this line lay 
off a distance BP equal to 4 units of length, 
and P will be the point required. 

So, \ix — a and y— b, measure off AB equal to a units, and 
draw BP parallel to AY, and equal to b units. 

Hence, in order to determine the position of a point, we need 
only have the two equations 

x = a, y=b, 
in which a and b are given. These equations are therefore 
called the equations of a point. 

13. Signs of the co-ordinates. It is however necessary, in 
order to determine the position of a point, that not only the 
absolute values of a and b should be given, but also the signs of 
these quantities. If the axes are produced through the origin 
to X' and Y', it is obvious that the abscissas reckoned in the 
direction AX' ought not to have the same sign as those reck- 
oned in the opposite direction AX, nor should the ordinates 
measured in the direction AY' have the same sign as those 

B2 





34 ANALYTICAL GEOMETEY. 

measured in the opposite direction 
AY ; for if there were no distinction 
in this respect, the position of a point 
as determined by its equations would 
be ambiguous. Thus the equations of 
the point P would equally belong to 
the points P', P ;/ , P //; , provided the 
absolute lengths of the co-ordinates of these points were equal 
to those of P. This ambiguity is avoided by regarding the 
co-ordinates which are measured in one direction &s>j)lus, and 
those in the opposite direction as minus. It has been agreed 
to regard those abscissas which fall on the right of the axis 
YAY' as positive, and hence those which fall on the left must 
be considered negative. So also it has been agreed to consider 
those ordinates which are above the axis XAX' as positive, and 
hence those which fall below it must be considered negative. 

14. Equations of a point in each of the four angles. The 
angle YAX is called the frst angle ; YAX' the second angle; 
Y'AX' the third angle ; and YAX the fourth angle. 

The following, therefore, are the equations of a point in each 
of the four angles : 

For the point P in the first angle, x= +a, y— +l>. 

" P' " second angle, x = — a, y— + b. 

" P" " third angle, x=—a,y=—h. 

" P r// " fourth angle, x=+a y y=— b. 

If the point be situated on the axis AX, the equation y— b 

becomes y—0^ so that the equations 

x— ±0, y — 
denote a point in the axis of abscissas at the distance a from 
the origin. 

If the point be situated on the axis AY, the equation x~a 
becomes x = 0, so that the equations 

x = 0, y= ±b 
denote a point on the axis of ordinates at the distance b from 
the origin. 



CO-ORDINATES OF A TOINT. 



If the point be common to both axes, that is, if it be at the 
origin, its position will be denoted by the equations 

x = 0j y=0. 

The point P, whose co-ordinates are x, y, is often called the 
point (x, y); thus a point for which x = a,y=b is called the 
point (a, b). Hitherto the letters a and b have been supposed 
to stand for positive numbers, but they may also be used to 
represent negative numbers. 

Ex. 1. Indicate by a figure the position of the point whose 
equations are x— + 4, y= — 3. 

Ex. 2. Indicate by a figure the position of the point whose 
equations are a?= — 2, y— +7. 

Ex. 3. Indicate by a figure the position of the point 0, —5. 

Ex. 4. Indicate by a figure the position of the point —8, 0. 

Ex. 5. Indicate by a figure the position of the point —3,-2. 

Ex. 6. Draw a triangle the co-ordinates of w T hose angular 
points are 3,4; —3,-4; —1,0. 

15. Polar co-ordinates. The position of a point may also 
be denoted by means of the distance and direction of the pro- 
posed point from a given point. 

Thus, if A be a known point, and AX be 
a known direction, the position of the point 
P will be determined when we know the 
distance AP, and the ano;le PAX. 

Thus, if we denote the distance AP by ?', 
and the angle PAX by 0, the position of P is determined if r 
and are known. 

The assumed point A is called the pole ; the distance of P 
from A is called the radius vector ; the line AX is called the 
initial line ; and the radius vector, together with its angle of 
inclination to the initial line, are called the polar co-ordinates 
of the point. The point whose polar co-ordinates are r and 
is sometimes called the point r, 0. 



36 ANALYTICAL GEOMETRY. 

16. Unit for the measure of angles. The unit commonly 
employed in Trigonometry for measuring angles is the nine- 
tieth part of a right angle, called a degree / but a different unit 
is sometimes more convenient. Since angles at the centre of 
a circle are proportional to the arcs on which they stand, we 
may employ the arc to measure the angle which it subtends, 
and it is convenient to take as the unit of measure the arc 
which is equal to the radius of the circle. Since the circum- 
ference of a circle whose radius is unity is 27r, the measure of 
four right angles w T ill accordingly be 2tt ; the measure of one 

right angle will be ^; the measure of an angle of 45° will be 

7T 

4> etc. 

17. Negative values of polar co-ordinates. The position of 
any point might be expressed by positive values of the polar 
co-ordinates r and 0, since there is here no ambiguity corre- 
sponding to that arising from the four angles of the figure in 
Art. 13. It is, however, sometimes convenient to admit the use 

of negative angles, and in this case an an- 
gle XAP' is considered negative when it is 
measured in the direction corresponding to 
the mention of the hands of a watch ; and 
an angle is considered positive when it is 

measured in the opposite direction, as XAP. 

The same direction may be represented either by a negative 

angle or by a positive angle. Thus, if the angle XAP 7 be half 

7T 

a right angle, the direction AP' may be denoted either by — -j 

or +\. 

"We also sometimes admit negative as well as positive values 
of the radius vector. Thus, suppose the co-ordinates of P to 

7T 7T 

be a and -r ; that is, let XAP=-j, and AV — a; if we produce 
PA to P", making AP" = AP, then P" may be determined by 

7T 

saying that its co-ordinates arc — a and -r. The radius vector 




J 



CO-ORDINATES OF A POINT. 



37 



is considered positive when it is measured in the direction of 
the extremity of the arc measuring the variable angle ; it is 
considered negative when it is measured in the opposite direc- 
tion. 

Thus the co-ordinates 

7T 

j represent the point P. 



r and 


— r and 


— r and - 


— r and 


— r and 


— r and — 



7T 

2 

3ZH 
4 

7T 

4 
3* 



P 6 . 
P r 

P. 



v, 



p* 




p 7 
P 6 



Thus the same point P is denoted either by the co-ordinates 

r and -r« or — r and ^r, or — r and — 4~- 
4' 4 ' 4 

Ex. 1. Indicate by a figure the position of the point whose 

co-ordinates are a, 15°, where a~\ inch. 

Ex. 2. Indicate by a figure the position of the point 2a, 40°. 

Ex. 3. Indicate by a figure the position of the points 

Q7T ^777 7T 7T 

— a, 45°; —a,— 135°; 3&,^r-; 5a, -V; 2a sin. -^^-. 

18. Implicit equations of a point. The position of a point 
may be determined not only explicitly by co-ordinates, but im- 
plicitly by means of simultaneous equations which these co- 
ordinates satisfy. For if we have two simultaneous equations 
between two variables, we can find the values of these variables 
by the methods of Algebra, and these values are the co-ordi- 
nates of known points. 

Ex. 1. Thus, suppose we have the equations 
2x + 3y= 12, 
3x-2y=5, 
we find x = 3, and y=2. 



38 ANALYTICAL GEOMETRY. 

In tliis and the following examples the pupil should draw 
the figure representing the problem. 

Ex. 2. Determine the point whose co-ordinates satisfy the 
equations 5x— 4y=9, 

7x — 5y = 15. 

Ans. a? =5, and 2/ =4. 

Ex. 3. Determine the point whose co-ordinates satisfy the 
equations x y 

a b~ > 



x y n 
o a 



Ans. x — y— — — r. 



Ex. 4. Determine the points whose co-ordinates satisfy the 
equations x+y=4(x—y), ) 

Ans. (5, 3), and (-5, -3). 
Ex. 5. Determine the points whose co-ordinates satisfy the 

} 

Ans. (5, 3), (-5, -3), (4</2, -/2), and (-4^/2, - V2). 



equati ons x 2 + xy= 40, 1 



19. To find the distance of any point from the origin in 
terms of the co-ordinates of that point. 
Y| Cfe^ First. Let the co-ordinates be rectan- 

P gular. 

We have AP 2 = A B 2 + B P 2 = ^ 2 + y 1 ; 
therefore AP = -y/x* + y 2 . 



-^ - 1 * Ex. 1. Determine the distance from the ori- 

gin to the point whose co-ordinates are x=3a, y—^a. 

Ans. AP=V9a a +16a*=5a. 

Ex. 2. Determine the distance of the point —26, 3 J, from 
the origin. Ans. 1^13. 

Ex. 3. Determine the distance from the origin to the point 
a sin. /3, a cos./3. Ans. a. 




CO-ORDINATES OF A POINT. 39 

Ex. 4. Determine the distance of the point 5a, — 3$, from 
the origin. 

20. Case Second. When the co-ordinates are oblique. 
From P draw PD perpendicular to AX; 

then (Geom., B. IV., Prop. 13) 

AP 2 =AB 2 +BP 2 + 2AB m BD. 
But by Trig., Art. 41, 

R:cos.PBD::PB:BD. JT ~b" 

Hence BD=:PB cos. PBD (radius being unity). 

Therefore AP 2 = AB 2 + BP 2 + 2 AB . PB cos. PBD. 
But PBD= YAX, which we will represent by w. 

Hence AP = (x 2 +y 2 + 2xy cos. w) 2 . 

In the following examples we will suppose the axes to be 
inclined at an angle of 60°. 

Ex. 1. Determine the distance from the origin to the point 

3a,±a. Am AP = (9a 2 + 16a 2 + 24^ 2 xi)^:=aV37. 

Ex. 2. Determine the distance from the origin to the point 
-25,35. Arts. AP=5y7. 

Ex. 3. Determine the distance from the origin to the point 
a sin. ft a cos. 0. ^. a(l+i sin. 2/3)* 

' Note. Sin. 2A=2 sin. A cos. A (Trig., Art. 73). 

Ex. 4. Determine the distance from the origin to the point 
5a, —3a. 

21. To find the distance between two given points. 

Case First. Let the axes be rectangular. 

Let P and Q be the two points, and repre- 
sent the co-ordinates of P by x v y v and those 
of Q by x v y r 

Draw PR parallel to the axis of x, cutting a. m: n 
QN in E. 

Then PQ 2 =PE 2 +EQ 2 . 

But PR=MN=AN--AM=tf 2 -^, 

and QE = Q^_PM=2/ 2 -?/ r 



R 




40 ANALYTICAL GEOMETEY. 

Therefore PQ 2 = (x 2 -xtf + (y 2 -yj*, 

and PQ = V{x 2 -xtf + (y 2 -ytf. 

Ex. 1. Determine the distance between the point 3, 4, and 
the point 4, 3. Ans. PQ2 = (3-4) 2 +(4-3) 2 .\ PQ= V2. 

Ex. 2. Determine the distance between the point —3,4, and 
the point 4, —3. Ans. 7V%- 

Ex. 3. Determine the distance between the point 2, 2, and 
the point —2,-2. Ans. 4-/2. 

Ex. 4. Determine the distance between the point 2a, 0, and 
the point 0, —2a. Ans. 2a ^/ 2. 

Ex. 5. Determine the distance between the point — 2a, 2a, 
and the point 4$, —6a. 

22. Case Second. Let the axes be inclined at 
an angle w. 

Then, as in Art. 20, 
A ra & * PQ 2 =PK 2 +KQ 2 -f 2PK.KQ cos.YAX, 
or PQ = V(x 2 -x l ) 2 + (y 2 --y l ) 2 + 2(x 2 -x l )(y 2 -y 1 ) cos. w. 

Ex. 1. Determine the distance between the point 0, 3, and 
the point 4, 0. 

Ans. PQ 2 =4 2 +3 2 -2.4.3 cos. w = 25-24 cos. *>, 

and PQ = V25-24 cos. w. 

Ex, 2. Determine the distance between the point 0, 3, and 
the point -4, 0. Ans. V25 + 24 cos. w. 

Ex. 3. Determine the distance between the point 2, —2, and 
the point -2, 2. An*. 8 fa* 

Note. 2 sin. 2 A = 1 - cos. 2A (Trig., Art. 74). 

Ex. 4. Determine the distance between the point a, 0, and 

the point 0, a. A ~ . w 

1 ' Ans. 2a sin. -~- 

23. Case Third. Let the co-ordinates be polar. 

P Let P and Q be the two given points ; repre- 

>q sent AP by r v and AQ by r 2 ; also PAX by V 
"p and QAX by 2 . 

From P draw PD perpendicular to AQ. 




CO-ORDINATES OF A TOINT. 



41 



By Geom., Bk. IV., Prop. 12, 

PQ 2 = AP 2 + AQ 2 -2AQ x AD. 
But AD = AP cos. PAQ (radius being unity). 

Hence PQ 2 = AP 2 + AQ 2 -2AP x AQ x cos. PAQ 

=r 2 +r 2_2r x r 2 , cos. (0,-02), 
and PQ = Vr x 2 +r 2 2 — 2r x r 2 , cos. (0 1 — 9 2 ). 

Ex. 1. Determine the distance between the point 2a, 30°, and 
the point a, 60°. Ans. PQ 2 =4& 2 +a 2 -4a 2 x £V3, 

and PQ =aVo— 2-/3. 
Ex. 2. Determine the distance between the point a, 0°, and 
the point b, 30°. Ans. FQ 2 =a 2 +b 2 —2ab x £V3, 

and PQ = Va 2 +b 2 -abV'3. 
Ex. 3. Determine the distance between the point a, 0, and 
the point — a, — 0. 

Ans. FQ 2 =a 2 +a 2 + 2a 2 cos. 20=2a 2 (l + cos. 20), 
and PQ = 2a cos. 0. 
Note. 2 cos. 2 A = 1 + cos. 2 A (Trig., Art. 74). 
Ex. 4. Determine the distance between the point a, 0, and 
the point a,—Q. Ans. 2a sin. 0. 

24. To find the co-ordinates of the point which bisects the 
straight line joining two given points. 

Let D be the point required, AN, DN its co-ordinates, and 
let DN cut BF in E. 

Then 
AN"=AL+LN=»AL+BE=AL+iBF 

that is, AN=« l +^=S=5±S. 




A L 



In like manner, 



DN= 



2/1+2/2 



2 



Ex. 1. Determine the co-ordinates of the point of bisection 
of the line joining the point — 1, 1, with the point 3, —5. 

Ans. x—\, y——2. 

Ex. 2. Determine the co-ordinates of the point of bisection 
of the line joining the point 3, —3, with the point 5,-5. 




42 ANALYTICAL GEOMETEY. 

25. To find the area of a triangle whose angular points 
are given. 

Let BCD be the triangle, and let the co- 
ordinates of B, C, D be x x y v x 2 ?/ 2 , x 3 y 3 re- 
spectively. 
The area BCD 
l m n A = BCML + CDNM- BDNL. 
But BCML=iLM(BL+CM)=i(a? a -a 1 )(y 2 +y 1 ). 

So also CDNM = i(x 3 - a? 2 )(y 3 + y 2 \ 

and BDKL=£(# 3 — a^Xj^+jO. 

Therefore the area BCD = 

= XfWs + X &1 + X ^2 - X &2 - ^3 - ^ l)» 

Ex. 1. Determine the area of the triangle whose angular 
points are 3, 4 ; — 3, — 4 ; 0, 4. Ans. 12. 

Ex. 2. Determine the area of the triangle whose angular 

points are 0,0; 1,2; 2,1. 3 

jL±.ns, q. 

Ex. 3. Determine the area of the triangle whose angular 
points are a, ; —a, ; 0, b. Ans. ab. 

Ex. 4. Determine the area of the triangle whose angular 
points are 1,1; —1,2; —1,1. 

26. To convert the rectangular co-ordinates of a point into 
polar co-ordinates, and vice versa. 

Let x and y denote the co-ordinates of P referred to the rect- 
T angular axes AX and AY. Also, let r and 

denote the polar co-ordinates of P, the pole 
being at the origin A, and AX being the initial 
line. Draw PD perpendicular to AX. Then, 
by Trig., Art. 41, 

AD=AP cos. PAD, or x — r cos. 0; 
also PD = AP sin. PAD, or y — r sin. 0, 

which equations enable us to deduce the rectangular co-ordi- 
nates of a point from the polar co-ordinates. 



CO-OEDINATES OF A POINT. 43 

Again, AD 2 -f PD 2 =AP 2 , or tf + 7 f= r ^ 

and AD : E : : PD : tang. PAD, or - = tang. 0, 

which equations enable us to deduce the polar co-ordinates of 
a point from the rectangular co-ordinates. 

Ex. 1. Find the polar co-ordinates of the point whose rect- 
angular co-ordinates are 8=1, y=l, and indicate the point by 
a figure. Ans. r= i/2, 0=45°. 

Ex. 2. Find the polar co-ordinates of the points whose rect- 
• angular co-ordinates are 

(1) *=-l, y= + 2. 

(2) *=-!, y=-a. 

(3) x=+l, y=-2. 

Ex. 3. Find the rectangular co-ordinates of the point whose 

7T 

polar co-ordinates are ^=3, 0=o- A 3 3 ,_ 

rf J.7W.«=2,y=2V3. 

Ex. 4. Find the rectangular co-ordinates of the points whose 
polar co-ordinates are 

(l)r=+3, 0=-| 
(2)r=-3, 0=+|. 
(3) r=-3, 0=-f. 



44 



ANALYTICAL GEOMETRY. 



SECTION II. 



THE STKAIGHT LINE. 



28. Definition. The equation of a line is the equation 
which expresses the relation between the two co-ordinates of 
every point of that line. 

Hence, if any point be taken upon the line, and the values 
of x and y for that point be substituted in the equation, the 
equation will be satisfied ; and conversely, if the co-ordinates 
of any point whatever of the plane satisfy the equation of a 
line, that point will be on the line. 




29. To find the equation to a straight line referred to rect- 
angular axes. 

Let A be the origin of co-ordinates, 

AX and AY be rectangular axes, and 

let PC be any straight line whose 

equation is required to be determined. 

Take any point P in the given line, 

and draw PB parallel to AY; then 

will PB be the ordinate and AB the abscissa of the point P. 

From A draw AD parallel to CP, meeting the line BP in D. 

Let AB=a?, 

BP=y, 
tang. PEX or DAX=m, 
and AC or DP =c. 

Then, by Trigonometry, Theorem IL, Art. 42, 
AB : BD : : radius : tang. DAX ; 
that is, x : BD : : 1 : m, 

or T>D=nix. 

But BP = BD + DP; 

that is, y = mx+c. 



THE STRAIGHT LINE. 



45 



Hence the equation to a straight line referred to rectangular 
axes is y—mx-\-c; 

where x and y are the co-ordinates of any point of the line, m 
represents the tangent of the angle which the line makes with 
the axis of abscissas, and c the distance from the origin at 
which it intersects the axis of ordinates. 



30. Signs qfm and c. 

If the line CP cuts the axis of ordi- 
nates below the origin, then c or AC will 
be negative. 

In that case, BP=BD-DP ; 



or, 



y = mx — c. 





The ano-le which the line makes with 
the axis of abscissas is supposed to be measured from the axis 
XX around the circle by the left. If the 
line CP is directed downward toward the 
right, as in the annexed figure, the line 
makes either an obtuse angle, CEX, with 
the axis of abscissas, or the negative acute 
angle CEA, the tangent of either of which 
angles is negative (Trig., Art, 69). 

In this case we have 

AB : BD : : radius : tang. DAX, 
or x : BD : : 1 : m. 

The tangent of DAX being negative, BD is also negative. 

But BP=-BD+DP, 

and the equation becomes y— —mx+c, 

where it must be observed that the minus sign applies only to 
the quantity m, and not to x, for the sign of x depends upon its 
direction from the origin A. 

If the line CP is directed downward toward 

the right, and cuts the axis of ordinates below 

the origin, then c is negative as well as m; and 

since BP= — BD — DP, the equation becomes 

y—— nix— c. 




46 



ANALYTICAL GEOMETRY. 



It is to be remembered that the symbols x, y, m, and c may 
stand for negative numbers, and therefore the single equation 

y=zmx + c 
may represent any line whatever. 






31. Four different positions of a line. There may, there- 
fore, be four positions of the proposed line, and these positions 
are indicated by the signs of m and c in the general equation. 

1. Let the line cut the axis of X to the 
left of the origin, and the axis of Y above 
it ; then m and c are both positive, and 
the equation is 

y—-\-mx+c. 

2. If the line cuts the axis of X to the 
right of the origin, and the axis of Y be- 
low it, then m will still be positive, but 
c will be negative, and the equation be- 
comes y = + w*# — c- 

3. If the line cuts the axis of X to the 
right of the origin, and the axis of Y 
above it, then m becomes negative and c 
positive. In this case, therefore, the equa- 
tion is y = — mx + c. 

4. If the line cuts the axis of X to the 
left of the origin, and the axis of Y below 
it, then both m and c will be negative, so 
that the equation becomes 
y— — mx—c. 
If we suppose the straight line to pass 
through the origin A, then c will become 
zero, and the general equation becomes 

y^jnx, 
which is the equation of a straight line passing through the 




A 



p 

X 



THE STRAIGHT LINE. 47 

32. Direction of a line indicated. It will be seen that tlie 
direction of the proposed line is indicated by the symbol m. 
If m is very small and positive, the line whose 
equation is y — mx takes the position AP, 
near the axis AX. As m increases the line 
changes its position, and when m = l the line 
makes an angle of 45° with AX. As the 
value of m increases the line approaches AY, and coincides 
with it when m becomes infinite. 

If m is negative and very large, the line assumes the position 
AP", and as m decreases the line moves toward AX', and 
when m——\ the line bisects the angle YAX'. When m be- 
comes zero, the line coincides again with the axis of abscissas. 

So, also, if the point P is supposed to travel round A through 
the third and fourth quadrants, the value of m will be positive 
in the third quadrant and negative in the fourth. 

Ex. 1. Let it be required to draw the line whose equation is 

y = 2x+4:. 

Draw the co-ordinate axes AX, AY. 
Kow if in this equation we suppose x=0, 
the value of y will designate the point in 
which the line intersects the axis of ordi- 
nates, for this is the only point of the line 
whose abscissa is zero. This supposition 
will give y — ^- 

Hence, if we take AB = 4, B will be one point of the required 
line. 

Again, if in the proposed equation we suppose y~0 y the 
value of x which is found from the equation will designate 
the point in which the line intersects the axis of abscissas, for 
that is the only point of the line whose ordinate is zero. This 
supposition will give 

2a?=-4, 
or x=—2. 

Hence, if we lay off from A toward the left a distance AC 
= 2, C will be a second point of the proposed line. Draw the 




48 



ANALYTICAL GEOMETRY. 



straight line BC, and produce it indefinitely both ways ; it will 
be the line whose equation is y—2x+4^. 

The student should regard every algebraic equation in this 
treatise as expressing some geometrical truth, and he should 
accustom himself to express these truths in appropriate geo- 
metrical language. Thus the equation y — 2x +4 expresses the 
truth that the ordinate of a certain straight line is equal to 
twice the corresponding abscissa, increased by four. 

So also the general equation of a straight line, y=mx + c, 
expresses the truth that the ordinate of any straight line is 
equal to some multiple of the corresponding abscissa, in- 
creased by a constant number. 



33. Any number of points of a line determined. When the 
equation of a line is given, we may, if desired, determine any 
number of points of the line by assuming particular values for 
x, and computing the corresponding values of y. 

Thus, if in the equation ?/=2^+4 we suppose 
x — 1, we find y— 6. 







x—— 1, we find y= 2. 
y== 8. a=-2, " y=z 0. 

y=10. a?=-3, " y=-2. 

?/=12, etc. a?=— 4, " y~— 4, etc. 
In order to represent all these values by a 
figure, set off on the axis of abscissas lines 
equal to 1, 2, 3, etc., both to the right and left 
of A ; then erect a perpendicular from each 
of these points, and make it equal to the cor- 
responding value of y, setting it off above 
AX if the ordinate be positive, but below 
AX if negative. The required line must 
pass through all the points thus determined. 



34. Variables and constants. In the equation y=mx+c, 
m and c remain unchanged so long as we consider the same 
Straight line ; they are therefore called constant quantities, or 
constants. But x and y may have an indefinite number of 



THE STRAIGHT LINE. 49 

values, since we may ascribe to one of them, as x, any value 
we please, and find from the equation the corresponding value 
of y. x and y are therefore called variable quantities, or vari- 
ables. 

35. Meaning of the equation of a line. The equation of a 
line may be regarded as a statement of some geometrical prop- 
osition respecting that line. 

Thus the equation 

y=2x + l0 
may be regarded as the algebraic statement of the proposition, 
the ordinate of a certain line is always equal to twice its cor- 
responding abscissa increased by ten. 

36. Equation to a line parcdlel to one of the axes. If in the 
equation y=mx-\-c we suppose m=0, the line will be parallel 
to the axis of X, and the equation becomes 

y=0.x+c, 
or y— c. 

This is then the equation of a line parallel to the axis of X. 
If c is positive, the line is above the axis of X ; if negative, it 
is below it. 

So also x— dta is the equation to a straight line parallel to 
the axis of Y. 

Examples. Construct the lines of which the following are 
the equations : 

l.y=2x+3. 4:.y=—2x—5. 7. y=5. 

2. y=3x-7. 5. y=3x. 8. y=-2. 

3.y=—x+2. 6. y—x. 9.y=—x. 

37. Every equation of the first degree containing two vari- 
ables represents a straight line. 

Every equation of the first degree containing two variables 
can be reduced to the form 

Ax+By+C = Q, 
in which A, B, and C may be positive or negative. We shall 

C 




50 ANALYTICAL GEOMETRY. 

now prove that every equation of this form represents a straight 

line. 

Y ■ In this equation put y=0, and we have 


j<g. x=—-r, which represents the point D 

where the line intersects the axis of X. 

C 
Again, put x=0 y and we have y=z—^ 

which represents the point E where the line intersects the axis 
of Y. "We have thus determined two points in the line which 
this equation represents. 

Let P be any other point of the line or curve represented by 
the given equation. "We are to prove that P is on the straight 
line passing through the points D and E. 

Since P is supposed to be on the line represented by the 
given equation, its co-ordinates must satisfy this equation ; and 
representing its co-ordinates by x and y, we shall have 

Ax+By+C=0, 

whence y= tt- =PR. 

C C C -C-Aa> .. . ._ 
JSow ~T : ~t5 :: — a"""^ : ^ > identically. 

But these several terms are equal to those of the proportion 

AD:AE::DR:PR; 

that is, PR is a fourth proportional to the three lines AD, AE, 

and DP; that is, P lies on the straight line joining D and E, 

and the equation Ax+By+C — represents that straight line. 

If either A, B, or C be negative, the same demonstration will 
apply with a slight change of the figure. 

This equation always represents some straight line, and may 
be made to represent any one by giving appropriate values to 
A, B, and C. 

If in this equation A = 0, then the line is parallel to the axis 
of x ; if B = 0, the line is parallel to the axis of y ; if C = 0, 
the line passes through the origin. 

Exam/pies. Draw the straight lines represented by the fol- 
lowing equations : 






THE STEAIGIIT LINE. 



51 



l.x+y+10 = 0. 

2. a?+y=10. 

3. x+y=0. 

4. 2x+3y=0. 

5. 4a+3y=l. 



6. 2y= 3or— 5. 

7. ?/=4— #. 

8. 2^=?/+7. 

9 -4-^-1 
J '2 + 3 ' 

10.y-3 = 2(a>-2). 



11. # = 2y. 

12. X = 4:. 

13.y=2. 

14. 4#— 3y=l. 

15. #— 2y=— 4. 



38. 76> ^tm? tfAg equation to a straight line which passes 
through a given point. 

When a point P is not completely determined, its co-ordi- 
nates are denoted by the variables x and y ; but when the po- 
sition of a point is completely known, the co-ordinates are gen- 
erally denoted by the letters a, b, or by x y y y with suffixes, as x v 
Vv x v V2 5 or ^ x an d V with accents, as x', y\ x", y ;/ , etc. 

Let PCE be the straight line, C the given 
point whose co-ordinates are x v y v and P 
any point of the line whose co-ordinates are 
x and y. Draw the ordinates CL, PM ; also 
draw CD parallel to AX. 

Now PD=y-y 1 , 

and CD=#— x y 

But CD : PD : : radius : tang. PCD. 

Hence 




CD" 



tang. PCD, which we will represent by m. 
' or y-y^mix-x,), 



VU ~~~tA/-> 



■m. 



That is, 

which is the equation of a straight line passing through a given 
point P. 

Since the coefficient m y which fixes the direction of the line, 
is not determined, there may be an infinite number of straight 
lines drawn through a given point. This is also apparent from 
the figure. • 

39. Line passing through a given point and parallel to a 
given line. If it be required that the line shall pass through 
a given point, and make a given angle with the axis of X, then 



52 ANALYTICAL GEOMETRY. 

m becomes a known quantity, and if we put m' for the tangent 
of the given angle we shall have 

which is the equation of a straight line passing through a giv- 
en point, and making a given angle with the axis of X. 

Ex. 1. Draw a line through the point w T hose abscissa is 5 and 
ordinate 3, making an angle with the axis of abscissas whose 
tangent is equal to 2, and give the equation of the line. 

Ans. The equation is y — 2# + 7 = 0. 

Ex. 2. Find the equation to the straight line which passes 
through the point (a, &), and makes an angle of 30° with the 
axis of X. Ans. x—a — (y—b)-\/3. 

Ex. 3. Find the equation to the line which passes through 
the point (4, 4), and makes an angle of 45° with the axis of X. 

40. To find the equation to the straight line which jxesses 
through two given points. 

Let B and C be the two given points, 
the co-ordinates of B being x x and y v and 
the co-ordinates of C being x 2 and y 2 . 
Then, since the general equation for ev- 
ery point in the required line is 

y=mx + c, (1) 

it follows that when the variable abscissa x becomes x v then y 
will become y l ; hence 
• y l = mx 1 + c. (2) 

Also, when the variable abscissa x becomes x 2 , then y be- 
comes ?/ 2 , and hence y 2 =mx 2 +c. (3) 

By combining these three equations we may eliminate m 
and <\ 

If we subtract equation (2) from equation (1), we obtain 

y— ftsf^B-q). (4) 

Also, if we subtract equation (3) from equation (2), we ob- 
tain yi-Vz= m ^\~ x ^ 

from which we iind m—— — — . 




THE STRAIGHT LINE. 



53 



Substituting this value of m in equation (4), we liavc 



y-Vi=, 



V\-y* 



(x-x x \ 



which is the equation of the line passing through the two given 
points B and C. 

It is evident from the figure that — — ~ denotes the tangent 
of the angle BCD or BEX. ^ l ~^ 2 

If the origin be one of the proposed points (# 2 , y 2 ), then x 2 
= and y 2 — 0, and the equation becomes 



y~- 



x, 



-x, 



which is the equation to a straight line passing through the or- 
igin and through a given point. 

Ex. 1. Find the equation to the straight line which passes 
through the two points whose co-ordinat.es are a?, =7, ^==4, 
and # 2 =: 5, y 2 =3, and determine the angle which it makes 
with the axis of abscissas. 

Ex. 2. Find the equation to the straight line which passes 
through the two points x l = 2, ^ = 3, and x 2 — k, y 2 z = 1 5- 

Ex. 3. Find the equations to the straight lines which pass 
through the following pairs of points : 



(1) 


*i=3,y,=4; 


and 


*2 =1 J Vi = % 


(2) 


a?, = 5, y 1 = 6 ; 


u 


^2=- 1 , 9 J2 = - 


(3) 


^X^ 1 , Vx = 2 '> 


a 


<V=2,y 2 =— *• 


(4) 


*i=±>yi=-2; 


u 


» 2 =-3, y 2 =-5 


(5) 


^,=3,Vt=-2; 


a 


«r=0> 2/ 2 =°- 


(6) 


aJ i= 2 >yi= 5 > 


u 


a; 2 = 5 y 2 =-^ 


(7) 


»i=°>yi=i; 


a 


* 2 =l>y 2 =-l- 


(8) 


x i=°> Vi=— «; 


a 


* 2 =0,y 2 =— *. 


(9) 


x x =a,y,=b; 


(( 


a " 2 =^2/ 2 =-^ 


(10) 


a? |= a, y 1= _J; 


u 


*V=— <*,y 2 =-& 



41. Definition. The distance from the origin to the point 
where a line intersects the axis of X is called the intercept on 
the axis of X ; and the distance from the origin to the point 



54 



ANALYTICAL GEOMETRY. 




where a line intersects the axis of Y is call- 
ed the intercept on the axis of Y. 

Thus, in the annexed figure, AB and AC 
are the intercepts of the line PC on the two 
axes. 



42. To find the equation to a straight line in terms of its 
intercepts on the two axes. 

Let B and C be the points where the 
straight line meets the axes of y and x 
respectively. Suppose AC=;<2, and AB 
= b. Let P be any point in the line, and 
: ^T\ X let x and y be its co-ordinates. Draw 
PD parallel to AY. Then, by similar 




triangles, v/e have 

that is, 
whence 



AB:DP 
b:y 



AC:DC; 
a\a— x, 



x y . 



which is the equation to a straight line in terms of its inter- 
cepts a and b. 

Ex. 1. Find the equation to a straight line which cuts off in- 
tercepts on the axes of x and y equal to 3 and —5 respectively. 

Ex. 2. Find the equation to a straight line which cuts off the 
intercepts —4 and 2. 

43. To find the angle included between two given straight 
lines. 

Let BC and DE be any two lines in- 
tersecting each other in P. Let the 
equation to the line BC be 
y-m x x+c v 
and the equation to the line DE be 
y=m 2 x+c 2 ; 

then m x will be the tangent of the angle BCX, and m 2 the tan- 
gent of the angle DEX. Now, because PCX is the exterior 




. THE STKAIGIIT LINE. 55 

angle of the triangle PEC, it is equal to the sum of the angles 
CPE and PEC ; that is, the angle EPC is equal to the differ- 
ence of the angles PCX and PEX, or 

EPC=PCX-PEX, 
whence tang. EPC^tang. (PCX-PEX), 

which, by Trig., Art. 77, 

• tang. PCX-tang. PEX 
" 1 + tang. PCX x tang. PEX 
m 1 — m 2 



'1+mjnJ 

le tangent < 
two riven lines. 



which denotes the tangent of the angle included between the 



to* 



44. To determine the co-ordinates ofthejpoint of intersec- 
tion of two given straight lines. 

Let the equation to one line be 

y=m x x+c v (1) 

and the equation to the other 

y=m 2 x+c 2 . (2) 

Since the co-ordinates of every point on a line must satisfy 
its equation, the co-ordinates of the point through which both 
the lines pass will satisfy both equations ; we must, therefore, 
find the values of x and y from (1) and (2) regarded as simul- 
taneous equations. We thus obtain 

x=— — S and y=~ 1 — - — — K 
m 2 —m x * m 2 — m x y 

which are the co-ordinates of the point of intersection of the 
two lines. 

Ex.1. Find the angle included between the lines x+y=l 
and y=x+2; also find the co-ordinates of the point of inter- 
section. A ^ 1 3 

Ana. 90°, x=-^y=^. 

Ex. 2. Find the angle between the lines x+3y=l and x—2y 
= 1 ; also the co-ordinates of the point of intersection. 

Ans. 45°, x=l, y=0. 
Ex.3. Find the angle between the lines x+y\/3 = and 



5G ANALYTICAL GEOMETEY. 






x—y<\/ 3 = 2; also the co-ordinates of the point of intersec- 

Ans. 60°, x=l, y= — -%-. 

Ex. 4. Find the angle between the lines oy— #=:0 and 2a? + 2/ 
=1 ; also the co-ordinates of the point of intersection. 

Ans. 81° 52', »=£, y=|. 

Ex. 5. Find the angle between the lines 3y— 2a?+l = and 
3,i' — y=0; also the co-ordinates of the point of intersection. 

Ex. 6. Find the angle between the lines x + y— 3 = and 
pj-f y=2; also the co-ordinates of the point of intersection. 

45. To find the equation to the straight line which passes 
through a given point, and is perpendicular to a given 
straight line. 

Let x x y x be the co-ordinates of the given point, and 

y—mx+c 

the equation to the given line. The form of the equation to a 
line through {x x y^ (Art 3S) is 

The tangent of the angle between the two lines is (Art. 43) 

1 + mm^ 

If the angle of intersection of the two lines be a right angle, 

its tangent must be infinite, and the denominator l + ?n?n l must 

become zero, so that we must have 

1 
m,= . 

1 VI 

Hence the required equation is 

2/-?/i=--(<*-*i), 
which is the equation to the straight line passing through the 
point (x x yd) Mid perpendicular to the line y — mx-\-c. 

46. Condition of perpendicularity. We conclude from the 

x 
last article that V=— — +0, 

J m l 



THE STRAIGHT LINE. 



57 




represents a line perpendicular to the line 

y=mx + c. 
The condition by which two straight lines are shown to be 
at right angles to each other may also be determined as follows : 

Let BC be a given line, and let tang. 
BCX=?n. 

Let DE be perpendicular to BC, and 
let tang. DlEX=7n l ; then 

tang. DEX=-tang. DEA, 

= — cotang. BCA; 

that is, m 1 = — -. (Trig., Art. 28.) 

lib 

Hence we see that when two lines are perpendicular to each 
other ', the tangents of the angles vjhich they make vnth either 
axis are the reciprocals of each other ^ and have contrary signs. 

Ex. 1. Find the equation to the line which passes through 
the origin, and is perpendicular to the line x+y—2. 

Ans. y—x. 

Ex. 2. Find the equation to the line which passes through 
the point % 1 = 2 y y 1 = —4:, and is perpendicular to the line 3y 
+ 2x— 1 = 0. Ans. 2y=3x-14:. 

Ex. 3. Find the equation to the line which passes through 
the point (8, 4), and is perpendicular to the line w^hose equation 
is y=2x— 16. 

Ex. 4. Find the equation to the line which passes through 
the point (—1, 3), and is perpendicular to the line 3#+4y+2 
= 0. 



47. To find the perpendicular distance of a given point 
from a given straight line. 

Let P be the given point, whose co-ordi- 
nates are x x y v and let BC be the given straight 
line whose equation is 

y — mx + c. 
From P draw PD perpendicular to BC, 
and PM perpendicular to AX, cutting BC in 

C2 




58 ANALYTICAL GEOMETRY. 

E. Now, since the above equation applies to every point of 
BC, it must apply to E ; that is, 

E^l=mx l + c. 
The perpendicular - PD = PE sin. PED. 
But PE=PM-ME=y 1 -?waj 1 -(?, 

and sin. PED = sin. CEM = cos. ECM = ^t^tt = 

sec. ECM 

1. 1 



Vl + (tang. ECM) 2 Vl + m 2 

Therefore PD ^^l!? 

Vl + m 2 ' 

which equation expresses the distance from the given point 

(#1^1 ) *° ^ e gi ven straight line. 

If the point P be at the origin, then x x = 0, y x — 0, and we 

have PD= ~ C , 

Vl + ?n* 

which equation expresses the distance of the proposed line 

from the origin. 

Ex. 1. Find the perpendicular distance of the point 2, 3 from 
the line x~\-y—l. Ans. 2-\/2- 

Ex. 2. Find the distance of the point — 1, 3 from the line 

5x+fy+2 = 0. . 11 

Ans. -=-• 
5 

Ex. 3. Find the distance of the point 0, 1 from the line x—3y 

=1. , 2^/10 

Ans. — = — • 
5 

Ex. 4. Find the distance of the point 3, from the line 
:+tj=1. Ans, 



2-3— VI3* 

Ex. 5. Find the distance of the point 1,-2 from the line 

x+y— 3 = 0. Ans. 2^/2. 

Ex. 6. Find the distance of the origin of co-ordinates from 

x i/ G 

the line ^+^—1. Ans. -rn* 




THE STRAIGHT LINE. 59 

Ex. 7. Find the distance of the point 3, — 5 from the line 
2x-§y + 7=0. 

Ex. 8. Find the distance of the point 8, 4 from the line y—2x 
-16. 

48. To find the equation to a straight line referred to ob- 
lique axes. 

Let A be the origin of co-ordinates ; let 
AX, AY be the oblique axes, and let PC 
be any straight line whose equation is re- 
quired to be determined. Take any point 
P in the given line, and draw PB parallel 
to AY ; then will PB be the ordinate and £ a~ 
AB the abscissa of the point P. Through the origin draw a 
line AD parallel to CP, meeting the line BP in D. 

Denote the inclination of the axes by w, and the angle DAX 
by a. Since PB is parallel to AY, the angle ADB is equal to 
DAY; that is, equal to w—a. 

Let x, y be the co-ordinates of P, and represent AC or DP 
by c. 

Then, by Trig., Art 49, 

BD : AB : : sin. a : sin. (w — a). 

Hence BD = — — j— '- — r. 

Sin. (w — a) 
But BP=BD+DP. 

__ x sin. a 

Hence ?/=- — 7 -+<?_ 

J sin. (w — a) ? 

which is the equation to a straight line referred to oblique 
axes. 

If we put m for -= — -A r. the equation becomes 

r sin. (a)— ay u 

y=mx+c, 
which is of the same form as the equation referred to rectan- 
gular axes, Art. 29. The meaning of c is the same as in Art. 
29 ; but the factor m denotes the ratio of the sine of the incli- 
nation of the line to the axis of X, to the sine of its inclination 



60 ANALYTICAL GEOMETRY. 



' 




to the axis o£ Y. When the axes are at right angles to each 
other, m becomes the tangent of a. 

49. To find the polar equation to a straight line. 
Let BC be any straight line, and P any 

point in it. Let A be the pole, AX the in- 
itial line, and let AD be drawn from A 
perpendicular to BC. Let AD =p, the an- 
gle DAX=a, and let the polar co-ordinates 
of P be r, 9 ; then we shall have 
AD=APcos.PAD; 
that is, p=r cos. (0— a), 

or r—jp sec. (0— a), 

which is the polar equation to a straight line. 

If AD be taken for the initial line, then a=0, and the equa- ' 
tion becomes r=p sec. 0, 

which is the equation to a right line perpendicular to the ini- 
tial line. 

To trace a right line by its polar equation, we find its inter- 
cept on the initial line by making 6 = 0. Then from the pole 
as a centre, with a radius equal top, describe a circle, and 
draw a tangent to this circle from the point first determined ; 
this tangent line will be the line required. 

50. To find the polar equation to a line passing through 
the pole. 

Let x and y denote the co-ordinates of P re- 
ferred to rectangular axes ; also let r and 9 de- 
note the polar co-ordinates of P, the pole being 



d x at the origin A, and AX being the initial line. 
Then, as in Art. 26, x=r cos. 0, 
and y = r sin. 9. 

Substituting these values in the equation 

y=?7ix, 
we have r sin. 9=mr cos. 9 ; 

therefore tang. = m / 



THE STRAIGHT LINE. 61 

that is, 0=a constant, 

which is the polar equation to a straight line passing through 

the pole. 

Examples. Draw the straight lines represented by the equa- 
tions 



l.rcosYfl-jW. 4. 0= 

2. rcos.(0— J) =4. 
3.rcosYo-|J=8. 



7T 

3' 

5. e=| 

6. 0=0. 




51. The following examples are designed to show how the 
preceding principles may be applied to the solution of geomet- 
rical problems. 

To determine whether the perpendiculars drawn from the 
vertices of a triangle to the opposite sides meet in a point. 

Let ABC be any triangle, and let AE, BF, CD 
be perpendiculars from A, B, and C upon the op- 
posite sides. 

Let A be the origin of co-ordinates ; let AB be 
the axis of abscissas, and AY, perpendicular to -^ » "~B 
AB, the axis of ordinates. Let the co-ordinates of C be x l y v 
and those of B be a? 2 , 0. 

Now if the abscissa of the point where AE and BF intersect 
is equal to AD, the intersection of these lines must be on CD. 
Since each of these lines passes through a given point and is 
perpendicular to a given line, its equation will be given by Art. 
45 ; but we must first find the equations to the lines AC, BC, 
to which they are perpendicular. 

Since AC passes through the origin and the given point C, 
its equation is (Art. 40) 

y=*fi (i) 

and since BF passes through a given point B(a? 2 , 0), and is 
perpendicular to (1), its equation is (Art. 45) 






62 * ANALYTICAL GEOMETRY. 



x 

y=-v@-**)- ( 2 ) 

Also, since BC passes through, the point B(# 2 , 0) and the 
point C^yJ, its equation is (Art. 40) 

1 2 

and since AE passes through the origin (0, 0), and is perpendic- 
ular to (3), its equation is 

J Vi W 

At the point where (2) and (4) intersect, their ordinates must 
be identical. Hence we may equate their values, and we have 

X^ X^ tZ/g 

2/i 2 ■ Vi ' 
whence x=x 1 ; 

that is, x, the abscissa of the intersection of BF, AE, is equal 

to x v the abscissa of the point C ; hence the perpendicular CD 

passes through that intersection, and the three perpendiculars 

meet in a point. 

52. To determine whether the three perpendiculars through 
the middle points of the sides of a triansgle meet in a point, 
y £ Let ABO be any triangle, and let D, E, F be the 

middle points of its sides. Let P be the point 
t^^e where two of the perpendiculars EP, FP meet ; 
now if the abscissa of P is equal to AD, the inter- 
B section of the lines EP, FP must be in the per- 
pendicular drawn from D. 

Represent the point C by (a?^), and the point B by (# 2 , 0). 

x u 
The co-ordinates of F are -^, ^ (Art. 24), and the co-ordi- 
nates of E are x,+x , y, 

Now the equation to AC, passing through the origin and the 
pointy, is |^ (1) 




THE STKAIGIIT LINE. 63 

and the equation to FP, which passes through F( -~ y ~ V and 
is perpendicular to (1), is ^ ' 

The equation to BC, passing through the point (x 2 , 0) and (a?^), 
is y=-^-{x-x 2 ), (3) 

(CC -4- CC 7/ \ 
1 q 2 , ^L 

At the point where (2) and (4) intersect, their ordinates must 
he identical ; and equating their values, we have 

1/ 1 l 1 2l 1 "" > 2 \ 

which gives x—~\ 

that is, #, the abscissa of the intersection of EP and FP, is equal 

03 

to ~, which is the abscissa of the point D ; hence the perpen- 
dicular from D passes through that intersection, and the three 
perpendiculars meet in a point. 



64 ANALYTICAL GEOMETRY. 






SECTION III. 

TRANSFORMATION OF CO-ORDINATES. 

53. When a line is represented by an equation with refer- 
ence to any system of axes, we can always transform that equa- 
tion into another which shall equally represent the line, but 
with reference to a new system of axes chosen at pleasure. 
This is called the transformation of co-ordinates, and may con- 
sist either in altering the relative position of the axes without 
changing the origin ; or changing the origin without disturb- 
ing the relative position of the axes ; or we may change both 
the direction of the axes and the position of the origin. 

54. To change the origin from one point to another without 
altering the direction of the axes. 

Let AX, AY be the primitive axes, and 
let A'X', A'Y' be the new axes, respective- 
ly parallel to the preceding. 

Let AB, A'B, the co-ordinates of the new 
origin referred to the old axes, be repre- 
A b M ^ sented by a and h ; let the co-ordinates of 

any point P referred to the primitive axes be x and y, and the 
co-ordinates of the same point referred to the new axes be x' 
and y f . Then we shall have 

AM=AB+BM=AB+A , M / , 
or x—a+x\ 

Also, PM = MM' + PM' = B A' + PM', 

or y=zb + y'. 

Hence, to find the equation to any line when the origin is 
changed, the new axes remaining parallel to the old, we must 
substitute in the equation to the line, a+x f for x, and b+y' 
for y. 




TRANSFORMATION OF CO-ORDINATES. G5 

These formulas are equally true for rectangular and oblique 
co-ordinates. 

Ex. 1. Find what the equation 2x + 3y=S becomes when the 
origin is transferred to a point whose co-ordinates are a = 3, 
J=l. Ans. 2x' + 32j'=-l. 

Ex.2. Find what the equation y+2x=z — 5 becomes when 
the origin is changed to the point (2, 1). 

Ans. y'+2x' = —10. 

Ex. 3. Find what the equation y=3x—7 becomes w T hen the 
origin is changed to the point (—2, —3). Ans. y' = Zx f —10. 

Ex.4. Find what the equation ?/ 2 + 4?/— 4# + 8 = becomes 
when the origin is changed to the point (1, —2). 

Ans. y*=4:X. 

55. To change the direction of the axes without changing 
the origin, both systems being rectangular. 

Let AX, AY be the primitive axes, and 
AX 7 , AY 7 be the new axes, both systems 
being rectangular. Let P be any point ; \ q . 

x, y its co-ordinates referred to the old 
axes ; x\ y f its co-ordinates referred to 
the new axes. Denote the angle XAX 7 A R N 

by 9. Through P draw PR parallel to AY, and PM parallel 
to AY 7 . From M draw MN parallel to AY, and MQ parallel 
to AX. 

Then aj=AE=AN-NE=AN-MQ. 

Also AN=AM cos. XAX 7 =z# 7 cos. 0, 

and MQ=PM sin. MPQ=2/ 7 sin. 9. 

Hence x—x' cos. 0— y' sin. 9. 

Also 2/=PK=QR+PQ=MN+PQ. 

But MN= AM sin. MAX=a?' sin. 9, 

and PQ=PM cos. NPQ=y' cos. 9. 

Hence y— 1 ^' sin. 9+y f cos. 9. 

Hence, to find the equation to any line when referred to 
the new axes, we must substitute in the equation to the line, 
x' cos. 9— y' sin. 9 for x, and x' sin. 9+y' cos. 9 for y. 




/* 



M 



-X 



66 



ANALYTICAL GEOMETKY. 



Ex.1. Find what the equation x+y=10 becomes when the 
axes are moved through an angle of 45°. 

V2 
Note. sin. 45° = cos. 45°=-y-. 

Here ®=|V2-|v 2 > 

By substitution, the given equation becomes x' = 5 V% Ans. 

Ex. 2. Find what the equation y=3x—6 becomes when the 
axes are moved through an angle of 45°. 

Ans. 2y'=x f —3V%- 

Ex. 3. Find what the equation ?/ 2 — x 2 =:6 becomes when the 

axes are moved through an angle of 45°. Ans. x'y' — Z. 

x ?/ 
Ex.4. Find what the equation o + o=l becomes when the 

axes are moved through an angle of 45°. 

56. To transform an equation from rectangular to oblique 

co-ordinates. 

Let AX, AY be the primitive axes, and 
AX', AY 7 be the new axes. Let P be any 
point ; x, y its co-ordinates referred to the 
old axes ; x f , y r its co-ordinates referred 
to the new axes. Through P draw PR 
parallel to AY, and PM parallel to AY 7 . 

Draw also MN parallel to AY, and MQ parallel to AX. De- 
note the angle XAX' by a, and the angle XAY' by j3. 
Then a=AR=AN+NR=AN+MQ. 

But AN = AM cos. X AX 7 = x' cos. a, 

and MQ=PM cos. PMQ=y / cos. ft. 

Hence x—x' cos. a + y f cos. ft. 

Also ?/=PR=QR+PQ=MN+PQ. 

But MN as AM sin. XAX' = x' sin. a, 

and PQ=PM sin. PMQ=y' sin. j3. 

Hence y—^' sin. a + y f sin. ft. 



Y' 


A 


P 


/ M 


/^ 


Q 


/^^^ 





N R 



TRANSFORMATION OF CO-ORDINATES. G7 

Hence, if we wish to pass from rectangular to oblique axes, we 
must substitute in the equation to the line, x f cos. a+y' cos. /3 
for x, and x' sin. a+y' sin. /3 for?/. 

If the origin be changed at the same time to a point whose 
co-ordinates referred to the primitive system are m and n, these 
equations will become 

x—m+x' cos. a+y' cos. /3. 
y=n+x' m sm. a+y' sin. [3. 

In the following examples the origin and the axis of X are 
supposed to remain unchanged. 

Ex. 1. Transform the equation y=4:—x from rectangular to 
oblique co-ordinates, the new axes being inclined to one anoth- 
er at an angle of 45°. Ans. #' + ?/y/2=4. 

Ex.2. Transform the equation y=3x from rectangular to 
oblique co-ordinates, the new axes being inclined to one anoth- 
er at an angle of 45°. Ans. 3x'+y\/2 = 0. 

Ex. 3. Transform the equation y—^—x from rectangular to 
oblique co-ordinates, the new axes being inclined to one anoth- 
er at an angle of 60°. Ans. y'(V3+l) + 2x'=8. 

Ex. 4. Transform the equation 2x=Sy +6 from rectangular 
to oblique co-ordinates, the new axes being inclined to one an- 
other at an angle of 60°. Ans. Zx' + y'(l-^S) = Q. 

57. To transform an equation from rectangular to polar 
co-ordinates. 

Let AX, AT be the rectangular axes ; let 
B be the pole ; and let BD, the initial line, 
be parallel to AX. 

Let P be any point ; a?, y its co-ordinates 

referred to the rectangular axes ; p, its po- J^ — ^ ^- 

lar co-ordinates. Draw PM, BC parallel to 
AY, and let a, b be the co-ordinates of B referred to the prim- 
itive axes. 

Now AM = AC + CM = AC + BD. 

But BD=BP cos. PBD=p cos. 0. 

Hence x=a+p cos. 0. 



u 



i> 



68 ANALYTICAL GEOMETKY. 

Also PM=DM+PD=BC+PD. 

But PD =BP sin. PBD = P sin. 0. 

Hence y—b+p sin. 0. 

Hence, to transform the equation to any line from rectangular 

to polar co-ordinates, we must substitute in the equation to the 

line, a+p cos. for x, and b+p sin. for y. 

In the following examples the pole is supposed to coincide 
with the origin, and the initial line with the axis of X. 

Ex. 1. Transform the equation x 2 + y 2 = 9 from rectangular to 
polar co-ordinates. Ans. p 2 (cos. 2 0+sin. 2 0) = 9, or p = 3. 

Ex.2. Transform the equation xy—4: from rectangular to 
polar co-ordinates. 

Note. Sin. 20=2 sin. cos. (Trig., Art. 73). 

Ans. p 2 sin. 20=8. 

Ex. 3. Transform the equation x 2 +y 2 = ?nx from rectangular 
to polar co-ordinates. Ans. p—m cos. 9. 

Ex. 4. Transform the- equation x 2 —y 2 = 3 from rectangular 
to polar co-ordinates. 

Note. Cos. 20= cos. 2 0-sin. 2 (Trig., Art. 73). 

Ans. p 2 cos. 20=3. 

58. To transform an equation from oblique to rectangular 
axes, find the values of x' and y f from the formulas of Art. 56. 

To transform an equation from polar to rectangular co-ordi- 
nates, deduce the values of p and from the equations of Art. 
57. These values are 

p > = (x-a) 2 +(y-b)% 

„ y—b 

and tang. 0=' . 

to x— a 



THE CIRCLE. 



GO 



SECTION IV. 

THE CIRCLE. 

59. Definition. A circle is a plane figure bounded by a 
line, all the points of which are equally distant from a point 
within called the centre. The line which bounds the circle is 
called its circumference. A radius of a circle is a straight 
line drawn from the centre of the circle to the circumference. 




60. To find the equation to a circle referred to rectangtdar 
axes when the origin of co-ordinates is at the centre. 

Let A be the centre of the circle, and 
P any point on its circumference. Let 
r be the raclius of the circle, and x, y the 
co-ordinates of P. Then, by Geom., Bk. 
IY.,Pr.ll, 

AB 2 +BP 2 = AP 2 ; 
or, x^+if — r 2 , 

which is the equation required. 

61. Points of intersection with the axes. If we wish to de- 
termine the points where the curve cuts the axis of X, we must 
put y=0, 

for this is the property of all points situated on the axis of ab- 
scissas. On this supposition, we have 

x— dtzr, 
which shows that the curve cuts the axis of abscissas in two 
points on different sides of the origin, and at a distance from 
it equal to the radius of the circle. 

To determine the points where the curve cuts the axis of or- 
dinate*, we make % = 0, and we find 

y=±r, 



70 



ANALYTICAL GEOMETRY. 



which shows that the curve cuts the axis of ordinates in two 
points on different sides of the origin, and at a distance from it 
equal to the radius of the circle. 

62. Curve traced through intermediate points. If we wish 
to trace the curve through the intermediate points, we reduce 
the equation to the form 

2/=± VV 2 — 3 2 , 
from which we may compute the value of y corresponding to 
any assumed value of x. 

Example. Trace the curve whose equation is x 2 +y* = 100. 

By assuming for x different values from to 11, etc., we ob- 
tain the corresponding values of y as given below. 



"When 3=0, y=z ±10. 
3=1,2/= ±9.95. 
3=2, y= ±9.80. 
x=3, 2/= ±9.54. 
»=4,y==9.16. 
3=5, y— ±8.66. 



When x= 6, y=±8.00. 

3= 7, ?/=±7.14. 

3 = 8, y= ±6.00. 

3= 9,j/= ±4.36. 

3 = 10, 2/= ±0.00. 

a?=ll, y is imaginary. 
When 3=0, ?/ will equal ±10, which 
gives two points, a and a 7 , one above 
and the other below the axis of X. 
When 3=1, y— ±9.95, which gives the 
points b and V. When 3=2, y— ±9.80, 
which gives the points c and c', etc. If 
we suppose x greater than 10, the value 
of y will be imaginary, which shows that 
the curve docs not extend from the centre beyond the value 

3 = 10. 

If 3 is negative, we shall in like manner obtain points in the 
third and fourth quadrants, and the curve will not extend to 
the left beyond the value x — —10. 

Since every value of x furnishes two equal values of y with 
contrary signs, it follows that the curve is symmetrical above 
and below the axis of X. 




THE CIRCLE. 



71 



63. To find the equation to a circle when the origin is on 
the circumference, and the axis oflLjpasses through the centre. 

Let the origin of co-ordinates be at A, a 
point on the circumference of the circle, and 
let the axis of X pass through the centre. 
Let r be the radius of the circle, and let x, y 
be the co-ordinates of P, any point on the 
circumference. Then CB will be represent- 
ed by #— r. 

Now CB 2 + BP 2 = CP 2 , 

or {x—ry + y 1 — r% 

whence y~ — 2rx— x*, 

which is the equation required. 




64. Points of intersection with the axes. If we wish to de- 
termine where the curve cuts the axis of X, we make y=Q y 
and we find x(2r— x) — 0. 

This equation is satisfied by supposing x~0 y or 2r— x=0, 
from the last of which equations we find x=2r. The curve, 
therefore, cuts the axis of abscissas in two points, one at the 
origin, and the other at a distance from it equal to 2r. 

To determine where the curve meets the axis of ordinates, 
we make x=0, which gives 

which shows that the curve meets the axis of ordinates in but 
one point, viz., the origin. 



65. Curve traced through intermediate points. In order to 
trace the curve through intermediate points, w r e reduce the 
equation to the form 

y— ±V%rx—x 2 , 
from which we may compute the value of y corresponding to 
any assumed value of x, as in Art. 62. 

Ex. 1. Trace the curve w T hose equation is y' 1 = 10x—x' 2 . 

By assuming for x different values from to 11, etc., we ob- 
tain the corresponding values of y as given on the next page. 



72 



ANALYTICAL GEOMETRY. 



When x=0, y=0. 

a? =2, 2/=4. 
aj=3, y=4.58. 
a? =4, y =4.90. 
a? =5, y=5. 



Whena= 6, y=4.90. 
#=7, 2/ =4.58. 
x— 8, 2/=4. 
as= 9, y=3. 
a?=10,y=0; 
#=11, y is imaginary. 



These values may be represented by a figure as in Art. 62. 
Ex. 2. Trace the circle x 2 +y 2 = 10y. 
Ex. 3. Trace the circle x 2 +y 2 = —10a?. 






66. To find the equation to the circle referred to any rect- 
angular axes. 

Let C be the centre of the circle, and 
P any point on its circumference. Let r 
be the radius of the circle ; a and b the 
co-ordinates of C ; x y y the co-ordinates 
of P. From C and P draw lines perpen- 
dicular to AX, and draw CD parallel to 
x AX. Then 
CD 2 +DP 2 =CP 2 ; 




B 



that is, 

which is the equation required. 

67. Varieties in the equation to the circle. If in the equa- 
tion (x—a) 2 +(y—b) 2 —r 2 we suppose a=0 and 5 = 0, the centre 
of the circle becomes the origin of co-ordinates, and the equa- 
tion becomes 

x 2 +y 2 =r 2 (asm Art. 60). 

If we suppose a=r and 5=0, the axis of X becomes a diam- 
eter, and the origin is at its extremity, and the equation be- 
comes (x — ry +y*— r 2 , 
whence y 2 = 2rx—x 2 (as in Art. 63). 

If we suppose a = and b — i\ the axis of Y becomes a diam- 
eter, and the origin is at its extremity, and the equation be- 
comes x 2 + (y — ?y = r* 
whence x 2 — 2ry — y 2 . 



THE CIRCLE. 73 

68. General equation to the circle. Expanding the general 
equation to the circle referred to rectangular axes, we have 

x 2 + y 2 -2ax-2by + a* + b 2 -r 2 = 0', 
and hence it appears that the general equation to the circle is 
of the form 

x 2 +y 2 +Ax+By+C = 0, 
where A, B, and C are constant quantities, any one or more of 
which in particular cases may be equal to zero. The equation 

A^ 2 + A?/ 2 +B^+C?/+Dr=0 
may be reduced to this form by dividing by A, and is therefore 
the most general form that the equation can assume when the 
co-ordinates are rectangular. 

69. To determine the circle represented by an equation. If 
we can reduce an equation to the form 

^+2/ 2 +A^+By + C = ? 

Mi TO 

we may determine the circle it represents ; for, adding — j — 

to both sides of the equation, and transposing C, we have 

f BY A 2 +B 2 „ 

By comparing this equation with that of Art. 66, we perceive 
that it represents a circle, the co-ordinates of whose centre are 

— -q, — 7T, and whose radius is 

If A 2 + B 2 <4C, the radius becomes imaginary, and the equa- 
tion can represent no real curve. 

Ex. 1. Determine the co-ordinates of the centre, and the ra- 
dius of the circle denoted by the equation # 2 -f y 2 + 4#— 8?/— 5 
= 0. 

This equation may be reduced to the form 
(^+2) 2 + (2/— 4) 2 = 25. 

Hence the co-ordinates of the centre are — 2, 4, and the ra- 
dius is 5. 

D 



(*+f) +(l 



74 ANALYTICAL GEOMETRY. 

Ex. 2. Determine the co-ordinates of the centre and the ra- 
dius of the circle denoted by the equation a? 2 +y 2 + 4?/-— 4#— 1 
= 0. Ans. Co-ordinates 2, — 2, radius 3. ' 

Ex. 3. Determine the co-ordinates of the centre and the ra- 
dius of the circle denoted by the equation x 2 + y 2 + Qx— 4?/— 36 
= 0. Ans. Co-ordinates — 3, 2, radius 7. 

Ex. 4. Determine the co-ordinates of the centre and the ra- 
dius of the circle denoted by the equation a? 2 +y 2 — 3x— 4?/ +4 
= 0. Ans. Co-ordinates -|, 2, radius -§. 

Ex. 5. Determine the co-ordinates of the centre and the ra- 
dius of the circle denoted by the equation x' 2 + y 2 — 2a(x—y] 



2 ] 

Ans. Co-ordinates a, —a, radius (2a 2 + e 2 ) 2 . 



: 



Ex. 6. Eind the equation to the circle whose radius is 9, and 
co-ordinates of the centre —1, 5. 

Ex. 7. Find the equation to the circle whose radius is 5a, and 
co-ordinates of the centre 3a, ^a. 

70. To find the polar equation to a circle when the origin is 
on the circumference, and the initial line is a diameter. 

Let A be the pole situated on the 
circumference of the circle ; let AX, 
passing through the centre, be the ini- 
tial line, and let P be any point on the 
circumference. Let r be the radius of 
the circle, and let p and 9 be the polar 
co-ordinates of P. 
The equation of the circle referred to rectangular axes (Art. 
63) is y 2 = 2ra— a? 2 . 

To transform this equation from rectangular to polar co-or- 
dinates (Art. 59), we must substitute for x, p cos. 0; and for 
?/, p sin. 0. 

Making this substitution, we obtain 

p 2 sin. *0=2rp cos. 0—p* cos. 2 0; 
or, by transposition, 

P 2 (sin. 2 + cos. 2 0) = 2rp cos. 0. 




THE CIRCLE. 75 

But sin. 2 + cos. 2 is equal to unity. 

Hence, dividing by p, we obtain 

p = 2r cos. 0, 
which is the polar equation of the circle. 

71. Points of the circle determined. When 0=0, cos. 0=1, 
and we have 

p = 2r=AB. 

As increases from to 90°, the radius vector determines 
all the points in the semi-circumference BPA ; and when 
= 90°, cos. = 0, and p becomes zero. 

From = 90° to = 180° the radius vector is negative, and is 
measured into the fourth quadrant, determining all the points 
in the semi-circumference below the axis of abscissas. From 
= 180° to = 360° the circumference is described a second 
time. 

Ex.1. The polar co-ordinates of P are jo = 10, 0=45° ; deter- 
mine the radius of the circle. 

Ex. 2. The radius of a circle is 5 inches, and p = 8 inches ; 
determine the value of 0. 

Ex. 3. The radius of a circle is 5 inches, and 0=60° ; deter- 
mine the radius vector. 

72. Definition. Let two points be taken on a curve, and a 
secant line be drawn through them ; let the first point remain 
fixed, while the second point moves on the curve toward the 
first until it coincides with it ; when the two points coincide, 
the secant line becomes a tangent to the curve. 

Suppose a straight line MP to intersect 
a curve in two points, M and P, and let T ^^ — T ' 
the line turn about the fixed point P until 
it comes into the position PM'. The sec- 
ond point of intersection, which at first was on the left of P, is 
now found on the right of P ; hence, in the movement of the 
straight line from the position MP to the position PM', there 
must have been one position in w T hich the point M coincided 




76 ANALYTICAL GEOMETRY. 

with P. In this position, represented by the line TT', the line 
is said to be a tangent to the curve. 

This definition of a tangent suggests a method of finding its 
equation which is applicable to all curves. 

73. To find the equation to the tangent at any jpoint of a 
circle. 

Let the equation to the circle be x 2 +y 2 =r 2 . 

Let x\ y f be the co-ordinates of the point on the circle at 
which the tangent is drawn, and x", y" the co-ordinates of an 
adjacent point on the circle. The equation to the secant line 
passing through the points a? 7 , y f and x'\ y" (Art. 40) is 

Now, since the points x', y' and x", y" are both on the cir- 
cumference of the circle, we must have 

"+tj"=r*=:x"*+!, 

y"*-y'* = x'*-x' r - 
y"-y> X " + X' 

wlience ti^' = -y^ 



x n +y n =r*=:x"*+y"\ 
or i/"-y n =x n -x m ) 



Substituting this value in equation (1), we obtain 

y-'!/=- S ^r^{x-x'), (2) 






which is the equation to the secant line passing through the 
two given points. 

Now when the point x\ y r coincides with the point x", y", we 
have &'=#", and y' — y" ; hence equation (2) becomes 

x' 

which is the equation to the tangent at the point x\ y', where x 
and y are the co-ordinates of any point of the tangent line. 
Clearing of fractions and transposing, we obtain 

or xx' + yi/'^r 2 , 

which is the simplest form of the equation to the tangent line. 



THE CIRCLE. 



77 



74. Points where the tangent cuts the axes. To determine 
the point in which the tangent inter- 
sects the axis of X, we make y=0, 
which gives 






or 



;/=AC, 




since x is AC when y=0. 

To determine the point in which 
the tangent intersects the axis of Y, we make x=0, which gives 



or 



J y 



Ex. 1. On a circle whose radius is 6 inches, a tangent line is 
drawn through the point whose ordinate is 4 inches ; determine 
where the tangent line meets the two axes ; also the angle 
which the tangent line makes with the axis of X. 

Ex. 2. Find the point on the circumference of a circle whose 
radius is 5 inches, from which, if a radius and a tangent line 
be drawn, they will form with the axis of X a triangle w T hose 
area is 35 inches. 



75. To find the length of the tangent drawn to the circle 
from a given point. 

Let P be a point without the circle from 
which a tangent line PM is drawn. Draw 
the radius AM, and join AP. Let the co-or- 
dinates of P be x, y. Then we have 

PM 2 = AP 2 -AM 2 . 
But AP 2 =£ 2 + 2/ 2 (Art.l9). 

Hence PM = (x 2 + tf-r 2 )*, 

which denotes the length of the tangent line from the point x y y. 
If x 2 + y 2 >r 2 , or the point P be without the circle, the tan- 
gent PM will be real ; if x 2 + y 2 =:i* 2 , or the point P be on the 
circle, the length of the tangent becomes zero ; if x 2 + y 2 <r 2 , or 




78 ANALYTICAL GEOMETRY. 

the point P be within the circle, the tangent is imaginary ; but 
the quantity r^—xf—y 1 represents the product of the segments 
of the chord drawn through P. 

Ex. 1. Find the length of the tangent drawn from the point 
-—7, +5, to a circle whose radius is 4. 

Ex. 2. Find the length of the tangent drawn from the point 
—3, —6, to a circle whose radius is 5. 

76. Definition. The normal at any point of a curve is a 
straight line drawn through that point perpendicular to the 
tangent to the curve at that point. 

77. To find the equation to the normal at any point of a 
circle. 

Let the equation to the circle be x*+y 2 =r 2 , and let x', y r be 
the co-ordinates of the point on the circle through which the 
normal is drawn. 

We have found (Art. 73, Eq. 3) that the equation to the tan- 
gent at the point x\ y' is 

y-y'=-??(*— «0j 

x f 
where — — denotes the tangent of the angle which the tangent 

if 

line makes with the axis of X. Hence (Art. 46) the equation 
to the norma! will be 

v f 

which, after reduction, becomes 

?/ 

and this is the equation to the normal passing through the giv- 
en point. 

?/ 

We have found (Art. 40) that y = '—x is the equation to a 

x 

straight line passing through the origin and through a given 

point ; hence the normal at any point of a circle passes through 

the centre. 



THE CIRCLE. 79 

78. To determine the co-ordinates of the points of intersec- 
tion of a straight line with a circle. 
Let the equation to the circle be 

x 2 +y 2 =r% (1) 

and the equation to the straight line be 

y=z?nx+c. (2) 

Since the co-ordinates of every point on a line must satisfy 
its equation, the co-ordinates of the points through which both 
of the given lines pass must satisfy both equations. We may 
therefore regard (1) and (2) as simultaneous equations contain- 
ing but two unknown quantities, and we may hence determine 
the values of x and y. By substitution in equation (1) we ob- 
tain 

x 2 + mV + 2cmx + c 2 = r 2 , 
or (1 + m 2 )x 2 + 2cmx =r 2 — c 2 , 

an equation of the second degree which may be solved by com- 
pleting the square. We thus find 

— cm ± Vr 2 (l + m 2 ) — & 

and since x has two values, we conclude that there will be two 
points of intersection. 

If r\l+m*) = c*, the two values of x become equal, and the 
straight line will touch the circle. If r 2 (l + ra 2 ) is less than e 2 , 
the straight line will not meet the circle. 

Ex. 1. Find the co-ordinates of the points in which the circle 
whose equation is x 2 + y 2 =:25 is intersected by the line whose 
equation is % + y=zl. * j #=:4, and y—— 3, 

" 1 or x=—3, and y — ^. 

Ex. 2. Find the co-ordinates of the points in which the circle 
whose equation is x 2 + y 2 = 25 is intersected by the line whose 
equation is x+y=5. * ( x = 5 y and y—0^ 

' ( or #=0, and ?/=5. 

Ex. 3. Find the co-ordinates of the points in which the circle 
whose equation is x 2 -\-y 2 = Go is intersected by the line whose 
equation is 3x + y =25. * ( a?=7, and y=4, 

or x — S y and ?/-!. 




80 ANALYTICAL GEOMETRY. 

Ex. 4. Find the points in which the line y=5x+2 intersects 
the circle y 2 +# 2 — 4?/— 13#=9. 

Ans.\ ®=Mndy=7, 

( or x— — f, an( l V— — ¥• 
Ex. 5. Find the points in which the line y = 3a? + 2 cuts the 
circle y 2 + x 2 — 4# + 4y = 7. 

79. To find the co-ordinates of the points of intersection of 
two circumferences. 

Let CPP', DPP' be two cir- 
cumferences which intersect in 
P and P'. Let A and B be the 
centres of the circles, r and r f 
their radii, and let AB, the dis- 
tance between their centres, be 
denoted by d. Assume the line 
AB as the axis of X, and let AY be drawn perpendicular to 
AX for the axis of Y. 

The equation to the circle CPP' is 

x 2 + y 2 = r\ (1) 

The equation to DPP 7 , the co-ordinates of whose centre are 
(d, 0) (Art. 66), is 

(x-d) 2 +y 2 =r'\ (2) 

Since the co-ordinates of every point of a circumference 
must satisfy the equation of the circle, the co-ordinates of the 
points through which both circumferences pass must satisfy 
both equations. We may therefore regard (1) and (2) as sim- 
ultaneous equations involving but two unknown quantities, and 
hence we may determine the values of x and y. Subtracting 
equation (2) from equation (1), we obtain 

2xd-d 2 = r 2 -r /2 , 

r*-. r ' 2 + d 2 
whence x — ^j . 

Substituting this value of x in equation (1), we have 

C r *-r" + d 2 V 

y= r -[ 2d \> 



THE CIRCLE. 81 

whence y = ± -^ V±dV-(r 2 -r" 2 + dJ, 

which gives the ordinates of the points of intersection of the 
two circles. 

The double sign of y shows that the two points of intersec- 
tion have the same abscissa AE, but two ordinates numerically 
the same and with contrary signs. Hence, when two circum- 
ferences cut each other, the line joining their centres is perpen- 
dicular to the common chord, and divides it into two equal 
parts. 

Ex. 1. Find the co-ordinates of the points of intersection of 
the two circumferences 

x 2 +y 2 = 25, and x' + tf + Ux^-lS. 

Ans. x= -2.714; y= ±4.199. 
Ex. 2. Find the co-ordinates of the points of intersection of 
the two circumferences x 2 +y 2 =:6, and x 2 + y 2 — §x= — 8. 

Ans. x=1.75; y= ±1.714. 
Ex. 3. Find the co-ordinates of the points of intersection of 
the two circumferences 

x 2 +y 2 — 2x— 4y=l 5 and x 2 + y 2 — 4:X— 6y= — 5. 

80. To find the equation to the straight line which passes 
through the points of intersection of two circles which cut each 
other. 

Let the equations of the two circumferences, whose centres 
are at B and C, be severally 

x 2 + y 2 + ax+by+c=0, (1) 

and x 2 + y 2 + a'x + b'y+c' = 0; (2) 

it is required to find the equation of the straight line passing 
through the points P and P' where these circumferences inter- 
sect. 

Since the co-ordinates of the points P and P' satisfy each of 
the above equations, we may treat them as simultaneous equa- 
tions containing two unknown quantities. 

Subtracting equation (2) from equation (1), we have 

(a-a')x+(b-b')y+c-c' = 0. (3) 

D2 



82 



ANALYTICAL GEOMETRY. 




Since this is an equation of the first 
degree between x and ?/, it is the equa- 
tion of a straight line (Art. 37) ; and 
since it must be satisfied by the co- 
ordinates of the two points P and P', 
it must be the equation of the straight 
line DE passing through those points, 
and is, therefore, the equation re- 
quired. 

If we combine equation (3) with the equation of either cir- 
cle, we shall obtain the values of the co-ordinates of the points 
of intersection as in Art. 79. 

In general, if we have any two equations of curves, and w T e 
add or subtract those equations as in the process of elimination 
in Algebra, we obtain a new equation, which is the equation of 
a new line or curve which passes through the points of inter- 
section of the first two curves. 



81. To find the equation to a circle which passes through 
three given points. 

We have found (Art 68) that the general equation to the 
circle is x 2 +y 2 + Ax+By+C=:0, 

where A, B, and C are constant for a given circle, but vary for 
different circles ; so that when A, B, and C are known, the cir- 
cle is fully determined. 

If the three points x'y', a?"y", x" f y" ! are on the circumfer- 
ence of a circle, the co-ordinates of each of these points must 
satisfy the equation of that circle. If then we substitute the 
values of x', y f in the general equation, we shall obtain an equa- 
tion which expresses the relation between the coefficients A, B, 
and C. So also, if we substitute successively the values of 
x n y" and x'"y"\ we shall obtain two other equations express- 
ing the relations between the same coefficients. We shall then 
have three simultaneous equations expressing the relations be- 
tween the three quantities A, B, and C, from which the values 
of these quantities can be determined. 



THE CIRCLE. 83 

Ex. 1. Find the equation to the circle which passes through 
the three points 1, 2 ; 1, 3 ; and 2, 5 ; also the co-ordinates of 
the centre and the radius of the circle. 

Substituting these values successively in the general equa- 
tion of the circle, we have 

A+2B + C + 5 = 0, 
A+3B + C + 10 = 0, 
2A + 5B + C+29 = 0, 
from which we find A=— 9; B= — 5; C = 14. 
Hence the equation to the circle is x 2 + y 2 — 9x—5y +14 = 0. 

Hence the co-ordinates of the centre are •§, -f ; and the radius 
is 1^/2. 

Ex. 2. Find the equation to the circle which passes through 
the three points 2, — 3 ; 3, — 4; and — 2, — 1 ; also the co-ordi- 
nates of the centre and the radius of the circle. 

Aiis. Eq., x 2 + y 2 + Sx + 20y + 31 = ; co-ordinates, - 4, - 10 ; 
radius = -\/85. 

Ex. 3. Find the equation to the circle which passes through 
the origin and through the points 2, 3 and 3, 4 ; also the co-or- 
dinates of the centre and radius of circle. 

Ans. Eq., x 2 + y 2 — 23x + lly = ; co-ordinates, -% 3 -, —^-; ra- 
dius =4<v/26. 

Ex. 4. Find the equation of the circle which passes through 
the three points — 4, — 4 ; — 4, — -2 ; — 2, +2 ; also the co-ordi- 
nates of the centre and radius of circle. 

Ans. Eq., x 2 + y 2 — 6x+6y— 32 = 0; co-ordinates, 3,-3; ra- 
dius, 5 -/2. 

Ex. 5. Find the equation of the circle which passes through 
the points —2, —4; 2, 2 ; 4, 4; also the co-ordinates of the 
centre and radius of circle. 

Am. Eq.,x 2 + y 2 —4:2x+30y+l6 = 0; co-ordinates, 21, -15 ; 
radius =5-y/26. 

Ex. 6. Find the equation of the circle which passes through 
the origin and cuts off lengths 6, 8 from the axes ; also the co- 
ordinates of the centre and radius of circle. 

Ans. Eq., x 2 + y 2 — Gx— 8y=0; co-ordinates, 3, 4 ; radius, 5. 



84 



ANALYTICAL GEOMETRY. 



SECTION V. 



THE PARABOLA. 




82. A parabola is a plane curve every point of which is 
equally distant from a fixed point and a fixed straight line. 

The fixed point is called the focus of the parabola, and the 
fixed straight line is called the directrix. 

Thus, if a straight line BC, and a point F 
without it be fixed in position, and the point 
P be supposed to move in such a manner 
that PF, its distance from the fixed point, is 
always equal to PD, its perpendicular dis- 
tance from the fixed line, the point P will 
describe a parabola of which F is the focus 
and EC the directrix. 

83. From the definition of a parabola the curve may be de- 
scribed mechanically by means of a ruler, a square, and a cord. 

Let BC be a ruler whose edge coincides 
with the directrix of the parabola, and let 
DEG be a square. Take a cord whose 
G length is equal to DG, and attach one ex- 
tremity of it at G and the other at the fo- 
cus F. Then slide the side of the square 
DE along the ruler BC, and at the same 
time keep the cord continually stretched by 
means of the point of a pencil, P, in contact 
with the square ; the pencil will trace out a portion of a parab- 
ola. For, in every position of the square, 
PF+PG=PD + PG, 
and hence PF = PD; 

that is, the point P is always equally distant from the focus F 
and the directrix BC. 




THE PAKAEOLA. 



85 



If the square be turned over, and moved on the other side 
of the point F, the other part of the same parabola may be de- 
scribed. 



84. A straight line drawn through the focus perpendicular 
to the directrix is called the axis of the parabola. The vertex 
of the axis is the point in which it intersects 
the curve. The chord drawn through the fo- 
cus of a parabola at right angles to the axis 
is called the latus rectum. 

Thus, in the figure, BX is the axis of the 
parabola, A is the vertex of the axis, and LI/ 
is the latus rectum. 




85. To find the equation to the parabola referred to rectan- 
gular axes. 

Take the directrix YY' as the axis of 
ordinates, and BX, drawn perpendicular to 
it through the focus, as the axis of abscis- 
sas. Let BF=2&. By the definition, 

FP=PD=BK 
Therefore FP 2 =BN 2 , 
or FN 2 +PN 2 =BN 2 ; 

that is, (x—2ay + y 2 =x' 2 , 

or y' 2 =4ca(x—a), 

which is the equation to the parabola. 

If in this equation we put y=0, we have x=a, which shows 
that the curve cuts the axis at a point A which bisects BF. 

The equation will be simplified if we put the origin at A. 
Let a?' = AN; then %=x' + a; and, since the axis of abscissas 
remains unchanged, y=y'. 

By substitution, equation (1) becomes 

y /2 = 4:ax f . 

We may suppress the accents if we remember that the origin 
is now at A ; thus we have 

jf = 4aa, (2) 




86 ANALYTICAL GEOMETKY. 

which is the equation to the parabola referred to its vertex as 
origin, and the axis of the parabola is the axis of X. 

86. To trace the form of the jparabola from its equation. 
Since y* = 4:ax y or x= j-, x can not be negative ; that is, the 

curve lies wholly on the positive side of the axis of y. 

Since y* = 4:ax, y=± 2(ax) 2 ; 

therefore, since this equation is unaltered if w y e write — y for 
y, to every point P on the curve on one side of the axis of X, 
there corresponds another point P' on the other side, such that 
P'JST^PN. Hence the curve is symmetrical with respect to 
the axis of X. 

Again, if x = 0, y=0, and has no other value ; therefore the 
curve does not meet either axis at any other point besides the 
origin. 

Also, the greater the value we give to x, the greater values we 
get for y ; and when x is infinite, y is infinite ; hence the curve 
goes off to an infinite distance on each side of the axis of X. 

87. To find the distance ofanyjpoint on the curve from the 
focus. 

The distance of any point on the curve from the focus is 
equal to the distance of the same point from the directrix. 
Hence FP = PD=BA+AN, 

or FP = & + #. 

88. To find the length of the latus rectum. 
In the equation 

y* = 4:ax, 
.put x—a; 

then y 1 = 4& 2 , 

and y=±2a, 

or the latus rectum LL/ ' — ka (see figure in Art. 84). 

If we convert the equation if — ^ax into a proportion, we 
shall have x:y::yAa ; 



THE PAKABOLA. 



87 



that is, the latus rectum is a third proportional to any abscissa 
and its corresponding ordinate. 

89. The squares of ordinates to the axis are to each other as 
their corresponding abscissas. 

Designate any two ordinates by y f , y'\ and the corresponding 
abscissas by x\ x" ; then we shall have 

y" = 4:ax\ 
y n * = ±ax". 
Hence y n : y ,n : : \.ax' : kax" : : x' : x n . 

Ex.1. The equation of a parabola is y 1 — ^. What is the 
abscissa corresponding to the ordinate 7 ? Ans. 12J. 

Ex. 2. The equation of a parabola is y 2 — lSx. What is the 
ordinate corresponding to the abscissa 7 ? 

Ans. ±Vl26. 
Ex. 3. The equation of a parabola is y*=zlOx. What is the 
ordinate corresponding to the abscissa 3 ? 

90. To trace the form of the parabola by means of points. 
If we reduce the equation of the parabola to the form 

y=:±2Vax, 
we may compute the values of y corresponding to any assumed 
value of x. 

Ex. 1. Trace the curve whose equation is y 2 = 4x. 
By assuming for x different values from to 5, etc., we ob- 
tain the corresponding values of y as given below. 
When x=+Q, y=0. 
" a?=l,y=±2. 
" x=2,y= ±2.828. 
« x=S,y=±3AG4:. 
" # = 4, y— ±4. 
" x =5,y= ±4.472. 
The first point (0, 0) is the origin ; the 
point (1, +2) is represented by a in the fig- 
ure ; the point (1, —2) by a f in the figure ; 
the point (2, +2.828) by b; the point (2, -2.828) by &', etc. 





88 ANALYTICAL GECXMETRY. 

Ex. 2. Trace the curve whose equation is y 2 — 18a?. 

Ex. 3. Trace the curve whose 
equation is x 2 = 9y. 

The curve will be of the form 
exhibited in the annexed figure, 
and is evidently a parabola whose 
axis is the axis of Y. 
Ex. 4. Trace the curve whose equation is y 1 — — 3x. 

91. To find the equation to the tangent at any jpoint of a 
parabola. 

Let the equation to the parabola be y^—^ax. 

Let x\ y' be the co-ordinates of the point on the curve at 
which the tangent is drawn, and a/', y' f the co-ordinates of an 
adjacent point on the curve. The equation to the secant line 
passing through the points x\ y' and #", y" (Art. 40) is 

y-y'=ijr~&-*')- W 

Now, since the points x' 9 y' and x", y" are both on the parabo- 
la, we must have y^—^ax^ 
and y"* = ±ax". 
Hence y'"-y"=:4.a(x"-x'), 

y n -y' ±a 
x // —x / ~y // + y r 
Substituting this value in equation (1), the equation of the 
secant line becomes 

4a 

y-y'=^7^>{x-xy (2) 

The secant will become a tangent when the two points coin- 
cide, in which case y'=zy". 
Equation (2) will then become 

2a 

y-y'=->(x-x% (3) 

which is the equation to a tangent at the point x\ y' . 
Clearing of fractions and transposing, w T e obtain 
yy' = 2a(x-x') + ?/'% 



or ,. 



THE PARABOLA. 



89 



yij = 2ax— 2ax r + 4# #', 
or yt/z=2a(x+x% 

which is the simplest iprm of the equation to the tangent line. 

92. Points where the tangent cuts the axes. To determine 
the point in which the tangent intersects 
the axis of X, we make y = , which gives 

= 2a(x+x'); 
that is, &= — x', 

or AT=-AR. 

To determine the point in which the 
tangent intersects the axis of Y, we 
make x — y which gives 

2ax f y" y f 




y- 



that is, 



y f ~2y'~~2' 
AB=iPR. 



93. Definition. A subtangent to a parabola is that part of 
the axis intercepted between a tangent and ordinate drawn to 
the point of contact. Thus TR is the subtangent correspond- 
ing to the tangent PT. 

From Art. 92 we see that the subtangent to the axis is bisect- 
ed by the curve. 

94. The preceding property enables us to draw a tangent to 
the curve through a given point. Let P be the given point ; 
from P draw PR perpendicular to the axis, and make AT = AR. 
Draw a line through P and T, and it will be a tangent to the 
parabola at P. 

95. To find the equation to a tangent to the parabola in 
terms of the tangent of the angle it makes with the axis. 

In the equation of a tangent line, 

2a 
2/-2/ / = ^-^ / )(Art.91,Eq.3), 

— represents the trigonometrical tangent of the angle which 

if 



90 ANALYTICAL GEOMETRY. 

the tangent line makes with the axis of the parabola (Art. 38). 

If we represent this tangent by m, we shall have 

2a _ y' a 

—=m, and 77 = —. (1) 

y ' . ' 2 m w 

The equation to a tangent line to the parabola (Art. 91) is 

yt/=2a(x+x% 

. 2a 2ax r 

whence ?/=— -,x+ — 7-, 

J y> T y , 

2& &ax' 



= —j x 



2a y f 
Hence, substituting equation (1), we have 

which is the equation to a tangent line. 
Hence the straight line whose equation is 

a 

y — mx + —. 

touches the parabola whose equation is y^ — ^ax. 

Ex. 1. Find the equation of a tangent to the parabola y' — X'&x 
at the point x' = 2 } y' = 6. 

Ex. 2. Find the equation, of a tangent to the parabola ?/ 2 :=4#, 
and parallel to the right line whose equation isy = 5x + l. 

Ex. 3. On a parabola whose equation is y*=zl0x, a tangent 
line is drawn through the point whose ordinate is 8. Deter- 
mine where the tangent line meets the two axes of reference. 

Ex. 4. On a parabola whose latus rectum is 10 inches, a tan- 
gent line is drawn through the point whose ordinate is 6 inch- 
es, the origin being at the vertex of the axis. Determine where 
the tangent line meets the two axes of reference. 

Ex. 5. Find the angle which the tangent line in the last ex- 
ample makes with the axis of X. 

Ex. 6. On a parabola whose latus rectum is 10 inches, find 
the point from which a tangent line must be drawn in order 
that it may make an angle of 35° with the axis of the parabola. 



THE PAKAB0LA. 



91 



96. Definitions. A normal is a line drawn through any 
point of the curve perpendicular to the tangent at that point, 
and terminated by the axis. 

A subnormal is the portion of the axis intercepted between 
the normal and the ordinate drawn from the same point of the 
curve. 



97. To find the equation to the normal at any point of a 
parabola. 

Let x\ y' be the co-ordinates of the given point. 

The equation to a straight line passing through this point 
(Art. 3S) is y—y'z=m(x—af) ; 

and, since this line must be perpendicular to the tangent whose 
equation is 

2/-2/'=!?(^-^)(Art91,Eq.3), 



V 



y 



we have m=z — — (Art. 45). 

Hence the equation to the normal is 



y-y=- 



y 

2d 



(x—x'). 



98. Point where the normal cuts the axis ofx. To find the 
point in which the normal intersects the axis of abscissas, make 
y=0 in the equation to the normal, 
and we have, after reduction, 

x—x' — 2a. 
But x is equal to the distance AN, 
and x f to AR; hence x—x f is equal 
to EN, which is equal to 2a; that is, 
the subnormal is constant, and is 
equal to half the Tatus rectum. 

Ex. 1. On a parabola whose latus rectum is 10 inches, a nor- 
mal line is drawn through the point whose ordinate is 6 inches. 
Determine where the normal line, if produced, meets the two 
axes of reference. 




92 



ANALYTICAL GEOMETRY. 



Ex. 2. Find the point on the curve of a parabola whose latus 
rectum is 10 inches, from which, if a tangent be drawn, and also 
an ordinate to the axis of X, they will form with the axis a tri- 
angle whose area is 36 inches. 

99. If a tangent to the parabola cuts the axis produced, the 
points of contact and intersection are equally distant from 
the focus. 

Let PT be a tangent to the parabola 
at P, and let PF be the radius vector 
drawn to the point of contact. 
We have found (Art. 92) 
TA=AK. 
Hence TF=AK+AF = FP (Art. 87); 
that is, the distance from the focus to 
the point where the tangent cuts the 
axis, is equal to the distance from the 
focus to the point where the tangent touches the curve. 




100. A tangent to the curve makes equal angles with the ra- 
dius vector and with a line drawn through the point of contact 
parallel to the axis. 

Let TT ; touch the parabola at P, and let BP be drawn 
through P parallel to AX ; then the angle BPT 7 is equal to the 
angle ATP. But since TF = PF,the angle FTP is equal to 
the angle FPT. Hence FPT is equal to BPT 7 , or the two lines 
FP and BP are equally inclined to the tangent. 

101. If a ray of light, proceeding in the direction BP, be in- 
cident on the parabola at P, it will be reflected to F on account 
of the equal angles BPT 7 and FPT. In like manner, all rays 
coming in a direction parallel to the axis, and incident on the 
curve, will converge to F. Also, if a portion of the curve 
revolves round its axis so as to form a hollow concave mirror, 
all rays from a distant luminous point in the direction of the 
axis will be concentrated in F. Thus, if a parabolic mirror be 




THE PARABOLA. 93 

held with its axis pointing to the sun, an intense heat and a 
brilliant light will be found at the focus. 

102. If from the focus of a parabola a straight line be 
drawn perpendicular to any tangent, it will intersect this tan- 
gent on the tangent at the vertex. 

Let the tangent PT be drawn, and from 
the focus F let FB be drawn perpendicu- 
lar to it ; the point B will fall on the axis 
AY, which touches the curve at A (Art. 
86). 

Since the triangle PFT is isosceles, the 
line FB, drawn perpendicular to the base 
PT, will pass through its middle point ; 
and since AT = AE (Art. 92), the line AY, which is parallel to 
PR, also passes through the middle point of PT ; that is, the 
line FB intersects PT in the same point with AY. 

Since the triangle FBT is right angled at B, we have 
FB 2 =FAxFT=FAxFP, 
or the perpendicular from the focus to any tangent is a mean 
proportional between the distances of the focus from the ver- 
tex and the point of contact. 

103. To determine the co-ordinates of the points of inter- 
section of a straight line with a parabola. 

Let the equation to the parabola be 

y'—^ax, (1) 

and the equation to the straight line be 

y=mx + c. (2) 

As in Art. 78, we may regard (1) and (2) as simultaneous 
equations, containing but two "unknown quantities. By substi- 
tution in equation (1), we obtain 

my 2 = 4:ay— kac. 
Completing the square, we obtain 

%a 2, a i 

y = — ± —(a — amc) * ; 
* m rn> J ' 



94 ANALYTICAL GEOMETRY. 

and, since y has two values, we conclude that there will be two 
points of intersection. 

If a—mc^ the two values of y become equal, and the straight 
line will touch the parabola. If a — mc is negative, the straight 
line will not meet the parabola. 

Ex. 1. Find the co-ordinates of the points in which the parab- 
ola whose equation is y*=4:X is intersected by the line whose 
equation is y — 2x— 5. 

Ans. 2/=4.3166, or -2.3166; # = 4.6583, or 1.3417. 

Ex. 2. Find the co-ordinates of the points in which the parab- 
ola whose equation is ?/ 2 = 18# is intersected by the line whose 
equation is y=2x— 5. 

Ans. 2/=12.5777, or -3.5777; #=8.7888, or 0.7111. 

Ex. 3. Find whether the parabola whose equation is y* = 16x 
is intersected by the line whose equation is y—2x + 2 y and, if 
there is a point of contact, determine its co-ordinates. 

Ex. 4. Find whether the parabola whose equation is y* = 16x 
is intersected by the line whose equation is y=2x+5. 

104. To determine the co-ordinates of the points of inter- 
section of a circle and parabola. 

If the centre of the circle is not restricted in position, there 
may be four points of intersection, corresponding to an equa- 
tion of the fourth degree, which can not generally be resolved 
by quadratics. If, however, the centre of the circle is upon 
the axis of the parabola, the several points of intersection may 
be easily found. 

Let the equation to the parabola be 

y* = 4:ax, 
and the equation to the circle be 

x* + tf=:r*; 
then, by substitution, we have 

and x=-2a±(4:a' i + ry, 

where x lias two values, but one of them is negative, and gives 

imaginary values for y. There will, therefore, be but two real 



THE PARABOLA. 95 

points of intersection. These have the same abscissa, and their 
ordinates will differ only in sign. 

Ex. 1. Find the co-ordinates of the points in which the parab- 
ola whose equation is if — 4:X is intersected by the circle whose 
equation is ^ 2 + ?/ 2 = 64. Ans. # = 6.2462; y= ±4.9985. 

Ex. 2. Find the co-ordinates of the points in which the parab- 
ola whose equation is y*=.l§x is intersected by the circle whose 
equation is x*+y*=:32x— 40. 

Anil'- 4 *?™* a /K 
\ y=±6-/2, or ±6^5. 

Construct the two curves, and show the points of intersec- 
tion. 

Ex. 3. Find the co-ordinates of the points in which the parab- 
ola whose equation is y* = 2x is intersected by the circle whose 
equation is x* + y* = 6x+5. 

105. To transform the equation to the parabola into anoth- 
er referred to oblique axes, and so that the equation shall pre- 
serve the same form. 

The formulas for passing from rectangular to oblique axes 
(Art. 56) are x—m-\-x' cos. a + y' cos. j3, 

y—n+x r sin. a + y' sin. fi. 

Substituting these values in the equation y^ — ^ax, and ar- 
ranging the terms, we have 

y /2 sin. 3 /3 + # /2 sin. 2 a + 2#y sin. a sin./3 + 
2(n sin. ft — 2a cos. fi)y' + n 2 —4cam = 2(2a cos. a— n sin. a)x', 
which is the equation to the parabola referred to any oblique 
axes. 

In order that this equation may be of the form y* = 4:ax, we 
must have the following conditions : 

1st. There must be no absolute term ; hence ?i 2 — 4:a?n = Q. 

2d. There must be no term containing a/ 2 ; hence sin. 2 a = 0. 

3d. There must be no term containing #'?/'; hence sin. a sin. 
0=0. 

4th. There must be no term containing y' '; hence n sin. /3 — 
2a cos. /3 = 0. 



96 



ANALYTICAL GEOMETRY. 



These equations contain four arbitrary constants, m, n, a, j3 / 
it is therefore possible to assign such values to the constants 
as to satisfy the four equations, and thus reduce the new equa- 
tion of the parabola to the proposed form. 

Since the equation y' — ^ax becomes ri* = 4cam by substituting 
the co-ordinates of the new origin for x and y, it follows that 
the first condition, ft 2 — 4$m = 0, requires the new origin to be 
on the curve. 

The second condition, sin. 2 a=0, requires the new axis ofx 
to he parallel to the axis of the parabola. 

The third condition, sin. a sin. j3 = 0, is satisfied by the sec- 
ond, without introducing any new condition. 

2a 2a 
Since —7 or — (Art. 95) has been found to represent the tan- 

gent of the angle which the tangent line makes with the axis 
of the parabola, the fourth condition, n sin. /3— 2a cos. /3 = 0, or 

— 75=:tang. j3= — , requires that the new axis ofj shall be 



cos 



tangent to the curve at the origin. 




If, therefore, the curve is 
referred to any tangent line 
A'Y', and a line A'X' drawn 
: ' through the point of contact 
parallel to the axis, the equa- 
tion becomes 

2/ /2 sin. 2 j3 = 
2{2a cos. a — n sm. a)x'; 
or, since sin. a = 0, 
and cos. a — 1, 



we have 



4<z 



sin. 2 j3 



If we represent -: — 775 by 4a', and omit the accents of the va- 
sm. ft 

riables, we shall have 

which is the equation required. 



THE TAEABOLA. 97 

106. Since the preceding problem furnished four arbitrary 
constants, ??z, n, a, /3, and required but three independent con- 
ditions (the second and third being but one), we may assign 
any value at pleasure to either of them ; that is, the new origin 
may be placed any where on the curve. 

107. From the equation of Art. 105, we find 

y— -±iV^a f x^ 
which shows that for every positive value of x there are two 
values of y equal numerically but having opposite signs, and 
these two values, taken together, form a chord PP 7 parallel to 
the axis of Y, which chord is bisected by the axis of X at P. 
So, also, the parallel chord QQ' is bisected by the axis of X at 
X. Hence a straight line parallel to the axis of the parabola 
bisects all chords parallel to the tangent at its extremity. 

108. Definition. A diameter of a parabola is a straight line 
drawn through any point of the curve parallel to the axis of 
the parabola. The vertex of the diameter is the point in which 
it meets the curve. 

109. The equation of Art. 105, y* = 4:a r x, is called the equa- 
tion of the parabola referred to a tangent line, and the diame- 
ter drawn through the point of contact ; and, since the new 
axis of Y is a tangent to the curve at the origin, a diameter bi- 
sects all chords parallel to the tangent at its extremity. 

The equation y* — ka'x 

shows that for oblique axes the squares of the ordinates are 
proportional to the corresponding abscissas, which is a gener- 
alization of the property proved in Art. 89. 

110. To determine the value of the coefficient of x in the 
equation y^ — ^a'x. 

From Art. 105 we have 

sin. fi 2a 

cos. /3 ~~ n ' 

E 



98 ANALYTIC Ah GEOMETRY. 

whence n sin. fi = 2a cos. /3, 

and n* sin. 2 j3=4a 2 cos. 2 j3, 

=4& 2 (l-sin. 2 /3), 
=4& 2 -4a 2 sin. 2 /3. 

Therefore sin. 2 /3 = ^ + 4( ^ - 

But from the equation to the curve referred to the original 
axes we have rt = 4#m / 

4a 2 a 

therefore Bin. P=^ m + w=j[+tf 

and -. — Ta=a + m = a. 

sin. p 

But m represents the abscissa of the 

new origin referred to the original axes ; 

hence 

a+m=FA'(Art. 87) = a', 

or the coefficient of x in the equation 

y 2 — 4:a f x is four times the distance from 

the focus to the new origin. 

111. To determine the length of the chord drawn through 
the focus parallel to the new axis of ordinates. 

If through the focus F the line BD be drawn parallel to the 
new axis of Y, then, calling x and y the co-ordinates of the 
point D, we have 

e=A'C=TF=A'F (Art. 99)= a 7 (Art. 110). 
But, by Art. 105, y* = ka'x ; 

hence ?/ 2 = 4taf xa' = ±a'% 

or y=%a\ 

and 2y—^a'\ 

that is, the coefficient Ota! is the double ordinate passing through 
the focus and corresponding to the diameter which passes 
through the origin. 

112. Definition. The parameter of any diameter is the 
double ordinate which passes through the focus. 




THE rAEAEOLA. 



99 



From Art. 110 we see that the parameter of any diameter 
is equal to four times the distance from the vertex of that di- 
ameter to the focus. 

In the equation y^ — ^a'x, ka' is the parameter of the diame- 
ter passing through the origin. 

The parameter to the axis is called the principal parameter, 
or latus rectum (Art. 84). 



113. To find the polar equation to the parabola, the focus 
being the pole. 

Let the axis be the initial line. Repre- 
sent FP by p , and PFN by 9. 
Tkcn PF = PD = BF + FN, 

or p — 2a + p cos. 9; 

I whence p(l — cos. 0)=:2tf, 

2a 

which is the polar equation to the parabola. 




114. If 0=180°, then cos. 0=— 1, and the value of p be- 
comes 

2a 

p=TTi 



=a=FA. 



If = 90°, then cos. 0=0, and the value of p becomes 

P = 2«=FC. 
If 0=0, then cos. 0=1, and we have 

2a 

the radius vector takes the direction AX, and does not meet 
the curve at a finite distance. 



115. If the variable angle be measured from the point A to- 
ward the right, then we must substitute for 9, 180° — 0, in which 
case cos. 0=— cos. 0', and we have 

2a 
^~l + cos. 0" 



100 



ANALYTICAL GEOMETRY. 



Ex. 1. What is the polar equation of a parabola whose latus 
rectum is 10, the pole being at the focus ; and what is the 
length of the radius vector for = 60° ? 

Ex. 2. The latus rectum of a parabola is 8 inches, and = 
135° ; what is the radius vector ? 

Ex. 3. The latus rectum of a parabola is 6 inches, and the 
radius vector is 10 inches ; determine the value of 0. 

Ex. 4. The radius vector of a parabola is 25 inches and 0~ 
135° ; what is the latus rectum ? 



M 






E^ 


M"' 




^<^ 






^^ 


P' 








P" 




A 


( 


P'" 




yR'" R" R' 


R 



116. To determine the area of a segment included between 
an arc of ct parabola and a chord perpendicular to the axis. 

Let PAQ be a segment of a parab- 
ola, bounded by the curve PAQ, and 
the chord PQ perpendicular to the 
axis AR. It is required to determine 
its area. 

Inscribe in the semi - parabola 
PAR a polygon PPT". . . . AR, and 
through the points P, P', P", etc., 
draw parallels to AR and PR, form- 
ing the interior rectangles P'R, P r/ R', 
etc., and the corresponding exterior rectangles P'M, P"M', etc. 
Designate the former by P, P', P", etc. ; the latter hyp,p', 
p", etc., and the corresponding co-ordinates by x, y, a?', y\ etc. 
We shall then have 

P / R=P / R / xRR / , 

V=y\x-xy 
FM=P / M , xffl , J 
p=x'(y-y'). 
P_ y(«-aQ ' 

p x \y-y ; y K ) 

But, since the points P, P', etc., are on the curve, we have 



N 



the rectangle 

or 

Also the rectangle 

or 



Whence 



if — ^ax^ y' 2 = 4(/,r' ; 



whence 



t// — Jb — 



i /a >i 

v -v __.-, „/ y 



Aa 



, and x' — -— . 
4# 



THE TAEAEOLA. 101 

Substituting these values in equation (1), we obtain 

r_ y / (?/-y /2 ) _y+?/_ 1 v 
p y'Xy-y) v' y'~ 

In the same manner we find 

r ; , y f 
jp' y m 

ii — 1+ ///, etc. 

If now we suppose the vertices of the polygons P, P', P ;/ , 
etc., to be so placed that the ordinates shall be in geometrical 
progression, we shall have 

y-tf-y! etc 

so that each interior rectangle has to its corresponding exterior 

y 

rectangle the ratio of 1 + — , to 1. 

y 

Therefore, by composition, 

P+F+P" + ,etc., y, 

j?+jp'+i>"+,etc., + y> 

that is, the sum of all the interior rectangles is to the sum of 

y 

all the exterior rectangles as 1 + — to 1. 

When the points P, P ; , P 7/ , etc., are taken indefinitely near, 

y 
the ratio — approaches indefinitely near to a ratio of equality ; 

the sum of the interior rectangles converges to the area of the 
interior parabolic segment APR, and the sum of the exterior 
rectangles to the area of the exterior parabolic segment AMP. 
Designating the former by S, and the latter by s, we have 

5=1+1=2, 

or S=2s#=f(S+s). 

But S + s is equal to the area of the rectangle AMPR; hence 
the parabolic segment APR is two thirds of the rectangle 
AMPR, or the segment PAQ is two thirds of the rectangle 
PMXQ. Hence the area of a parabolic segment cut off by a 



102 . ANALYTICAL GEOMETEY. 

double ordinate to the axis is two thirds of the circumscribing 
rectangle. 

Ex. 1. Determine the area of the parabolic segment cut off 
by a double ordinate whose length is 24 inches, the latus rect- 
um being 8 inches. 

Ex. 2. The area of a parabolic segment cut off by a double 
ordinate to the axis is 96, and the corresponding abscissa is 6. 
Determine the equation to the curve. 

117. By a demonstration like that of the preceding article, 
it may be also shown that the area of a parabolic segment cut 
off by the double ordinate of any diameter is two thirds of the 
circumscribing parallelogram . 

Example. Prove that if two tangents are drawn at the ex- 
tremities of any chord of a parabola, the segment cut off from 
the parabola is two thirds of the triangle formed by the chord 
and the two tangents. 



THE ELLIPSE. 103 




SECTION VI. 

THE ELLIPSE. 

118. An ellipse is a plane curve traced out by a point which 
moves in such a manner that the sum of its distances from two 
fixed points is always the same. The two fixed points are call- 
ed the foci of the ellipse. 

Thus, if F and F' are two fixed points, 
and if the point P moves about F in 
such a manner that the sum of its dis- 
tances from F and F ; is always the same, 
the point P will describe an ellipse, of 
which F and F' are the foci. The dis- 
tance of the point P from either focus 
is called the focal distance, or the radius vector. 

119. Description of the curve. From the definition of an 
ellipse the curve may be described mechanically. Thus, take 
a thread whose length is greater than the distance FF', and 
fasten one of its extremities at F, the other at F ; . Place the 
point of a pencil, P, against the thread, and slide it so as to 
keep the thread constantly stretched ; the point of the pencil 
will describe an ellipse. For in every position of P we shall 
have FP + FT equal to the fixed length of the thread ; that is, 
equal to a constant quantity. 

120. Definitions. The centre of the ellipse is the middle 
point of the straight line joining the foci. 

A diameter is any straight line passing through the centre, 
and terminated on both sides by the curve. 

The diameter which passes through the foci is called the 
transverse axis, or the major axis. 

The diameter which is perpendicular to the major axis is 
called the conjugate axis, or the minor axis. 



104 



ANALYTICAL GEOMETRY. 



The latus rectum is the chord drawn through one of the foci 
perpendicular to the major axis. 




121. To find the equation to the ellipse referred to its axes. 

Let F and F' be the foci, and 
draw the rectangular axes CX, 
CY, the origin, 0, being placed 
at the middle of FF'. Let P be 
-X any point of the curve, and draw 
PR perpendicular to CX. 

Let 2c denote FF 7 , the con- 
stant distance between the foci, 
and 2a denote FP + FT, the constant sum of the focal distan- 
ces. Denote FP by r, FT by r\ and let x and y denote the 
co-ordinates of the point P. 
Then, since 

FP 2 = PR 2 + RF 2 = PR 2 + (CR - CF) 2 , 
we have r 2 = y 2 + (x — c)\ (1) 

Also, PF /2 = PR 2 + RF /2 = PR 2 + (CR + CF)\ 
That is, r' 2 =y* + ( x + c)\ (2) 

Adding equations (1) and (2), we obtain 

r 2 + r' 2 = 2(y 2 + x 2 + c 2 ); (3) 

and subtracting equation (1) from (2), we obtain 

r /2 —r 2 = 4:Cx, 
which may be put under the form 

(r f + r)(r' —r) = kcx. (4) 

But from the definition of the ellipse we have 

r'+r = 2a. (5) 

Dividing equation (4) by equation (5), we obtain 

, 2cx 

r —r=- 



a 



Combining the last two equations, we find 



r=a- 



cx 
a 7 

ex 
a' 



(6) 

CO 



THE ELLIPSE. 105 

Squaring these values, and substituting them in equation (3), 
we obtain 

2 ^ ^ 2 i 2 i 2 

CO 

which may be reduced to the form 

a y + (a' - c> 2 = a\cC - &\ (8) 

which is the equation to the ellipse. 

This equation may, however, be put under a more convenient 
form. 

Represent the line BC by b. In the two right-angled trian- 
gles BCF, BCF', CF is equal to CF', and BC is common to 
both triangles ; hence BF is equal to BF 7 . But, by the defini- 
tion of the ellipse, BF+BF / = 2a; consequently BF = a. 
Now BC 2 = BF 2 -CF 2 ; 

that is, b' = a'-c\ (9) 

Substituting this value in equation (7), we obtain 

aY + b*tf=a*b\ (10) 

a? y Q 
or a* + b* = 1 > * (11 > 

which is the equation of the ellipse referred to its axes. 

This equation is sometimes written 

f=^-«n. (12) 

122. Points of intersection with the axes. To determine 
where the curve cuts the axis of X, make y — in the equation 
of the ellipse, and we obtain 

x—±a — CA or CA', 
which shows that the curve cuts the axis of abscissas in two 
points, A and A', at the same distance from the origin, the one 
being on the right, and the other on the left ;* and, since 2CA, 
or AA', is equal to 2a, it follows that the sum of the two lines 
drawn from any point of an ellipse to the foci is equal to the 
major axis. 

If we make x — in the equation of the ellipse, we obtain 

y=±b, = CBorCB', 

E2 



106 



ANALYTICAL GEOMETRY. 



which shows that the curve cuts the axis of Y in two points, 
B and B', at the same distance from the origin. 

123. Curve traced through intermediate points. If we w 7 ish 
to trace the curve through the intermediate points, we reduce 
the equation to the form 

b 



y=±-Vd 



a 



~ Jb i 



from which we may compute the value of y corresponding to 
any assumed value of x. 

Example. Trace the curve whose equation is 

49?/ 2 + 36^ = 1764. 
Solving the equation for ?/, we have 

6 , 

By assuming for x different values from to 7, we obtain 
the corresponding values of y as given below. 

When x=4,y= ±4.92. 
x=5,y=±4:.20. 
x = 6,y=+3.09. 
x=7, y=±0. 
When x — 0,y will equal ±G, 
which gives two points, a and a\ 
one above and the other below 
the axis of X. When x—l^y — 
± 5.99, which gives the points b 
and b '. When x — 2^ ?/=±5.75, 
which gives the points c and c\ 
etc. If we suppose x greater than 7, the value of y will be im- 
aginary, which shows that the curve does not extend from the 
centre be} r ond the value x = 7. 

If x is negative, we shall in like manner obtain points in the 
third and fourth quadrants, and the curve will not extend to 
the left beyond the value x= — 7- 

The ellipse is seen to be symmetrical above and below the 
axis of x, and also to the right and left of the axis of y. 



When^^O, y= ±6. 

x — 1, ?/=±5.94. 
x=2, y=±5.75. 
x=3, y=*5.42. 




THE ELLIPSE. 



107 



124. The circle is a particular case of the ellipse. When b 
is made equal to a, the equation of the ellipse becomes 

which is the equation of a circle ; hence the circle may he re- 
garded as an ellipse whose two axes are equal to each other. 



125. To find the foci of an ellipse when the ttoo axes are 
given. Since BF or BF' is equal to 
a (Art, 121), it follows that the dis- 
tance from either focus to the extrem- 
ity of the minor axis is equal to half A 
the major axis. 

If, then, from B, the extremity of 
the minor axis, with a radius equal to half the major axis, we 
describe an arc cutting the major axis A A' in F and F', the two 
points of intersection will be the foci of the ellipse. 




126. To find the length of the latus rectum. According to 
Art. 121, Eq. 12, 

b* 

tf = -la?-x 2 ). 



a 



Suppose x— c, 


or CF ; then 








v*=X&-<?\ 




where y 


is the 


ordinate at the point 


F. 


Eq.9, ' 




<t—e=&\ 




hence we 


have 






or 




a : b : : b : y, 




and 




2a :2b:: 2b: 2v. 





But by Art. 121, 



But 2y here represents the double ordinate drawn through 
the focus, and is called the latus rectum (Art. 120) ; hence the 
latus rectum of any ellipse is a third proportional to the ma- 
jor and minor axes. 



108 ANALYTICAL GEOMETRY. 

127. Equation of the ellipse in terms of the eccentricity. 

The fraction — , which represents the ratio of CF to CA, or the 
a 

distance from the centre to either focus, divided by half the 
major axis, is called the eccentricity of the ellipse. If we rep- 
resent the eccentricity by 0, then 

c 

- = e, or c=ae. 

But we have seen that & = c£—V L ; 
hence a 2 — 5 2 = aV, 

h% 1 » 

or —=z±—e. 

a" 

Making this substitution, the equation of the ellipse becomes 
which is the equation in terms of the eccentricity. 

128. To find the distance of any point on the curve from 

either focus. Equations (6) and (7) of Art. 121 are 

ex 
r=a+-, 

ex 

T—a—. 
a 

Substituting e for - these equations become 

r'=a+ex, 
r=a—ex, 

which equations represent the distance of any point on an el- 
lipse from either focus. 

Multiplying these values together, we obtain 
rr r — c^— <?V, 
which is the value of the product of the focal distances. 

The equation of an ellipse may assume forms differing from 
those of Art. 121, in consequence of multiplication or division 
by a constant, or of transposition. Thus, a? u + 4y 2 = 7; ?/ 3 = 25 
— 2x 2 ; ±(x' i + y' 2 ) = 7+x% are equations of ellipses referred to 
the centre and axes. 



THE ELLIPSE. 109 

Ex. 1. Trace the curve whose equation is 3^ 2 + 5?/ 2 =:15. 

Ex. 2. In a given ellipse, half the sum of the focal distances 
is 4, and half the distance between the foci is 3 ; what is the 
equation to the ellipse ? 

Ex. 3. In a given ellipse, the sum of the focal distances is 10, 
and the difference between the squares of half that sum and 
half the distance between the foci is 1G ; what is the equation 
to the ellipse ? 

Ex. 4. What is the eccentricity of the ellipse whose equation 
is9z 2 + 16?/ 2 = 144? 

Ex. 5. Trace the curve whose equation is x 2 + 4:y*=zl6. 

Ex. 6. Trace the curve whose equation is 3^ 2 + 4?/ 2 =:120. 

Ex. 7. What are the eccentricities of the ellipses of examples 
5 and 6 ? 

129, To find the equation of the ellipse when the origin is 
f at the vertex of the major axis. The 

equation of the ellipse when the ori- 
gin is at the centre is 

fj^-x\ (1) 

If the origin is placed at A', the or- 

dinates will have the same value as 

when the origin was at the centre, but the abscissas will be 

changed. If we represent the abscissas reckoned from A' by 

x', we shall have CR=A'R- A'C, 

or x=x'—a. 

Substituting this value of x in equation (1), we have 

which is the equation of the ellipse referred to the vertex A'. 

130. Iielation of ordinates to the major axis. If the last 
equation be resolved into a proportion, we shall have 

y 1 : (2a—x)x: :b*:a*. 
Xow 2a represents the major axis AA ; ; and since x repre- 




110 



ANALYTICAL GEOMETRY. 



sents A'E, 2a— x will represent AE; therefore (2a—x)x repre- 
sents the product of the segments into which the major axis is 
divided by the ordinate PE. Hence we have, the square of 
any ordinate to the major axis of an ellipse, is to the product 
of the segments into which it divides that axis, as the square 
of the minor axis, is to the square of the major axis. 

If we draw a second ordinate P'E' to the major axis, we 
shall have PE 2 : A'E x EA : : ¥ : a 2 : : P'E 72 : A'E' x E'A, 
or PE 2 : P'E' 2 ::A'Ex EA : A / E / x E'A ; 

that is, the squares of ordinates to the major axis of an ellipse 
are to each other as the products of the segments into which 
they divide that axis. 

131. Ordinates to the minor axis. The equation to the el- 
lipse, Art. 121, Eq. 10, may be put under the 
form 




or a* \b 2 \:x* \(b+y)(b-y). 

Now y represents CE; hence b + y repre- 
sents B'E, and b— y represents BE. Also x 
represents PE, which may be called an or- 
dinate to the minor axis. Hence we have the square of any 
ordinate to the minor axis of an ellipse, is to the product of 
the segments into which it divides that axis, as. the square of 
the major axis, is to the square of the minor axis. 

Example. The major axis of an ellipse is 12 inches, and the 
curve passes through the two points x = 4-, y — 0, and x~—^, 
y — 0) required the equation of the ellipse. 

132. An ordinate to the major axis of an ellipse is to the 
corresponding ordinate of the circumscribed circle, as the mi- 
nor axis is to the major axis. 

Let a circle be described on AA' as a diameter, and let the 
ordinate PE of the ellipse be produced to meet the circumfer- 
ence of the circle in P'. 



THE ELLIPSE. 



Ill 



The equation. of the ellipse when 
the origin is at the centre (Art. 121, 
Eq. 12) is 

^=^ , - ajS )=^- aj )(^+ a? )- A 

But {a— x)(a+x) represents ARx 
A'R ; hence we have 

PR 2 _V_ 
ARxA'R-V 
But P'R'=AR x A'R (Geom., B. IV., Pr. 23, Cor.) ; hence 

PR 2 V 




or 



P'R 9 ~a a 3 
FB:FB:;ft:a::2&:2fe 



133. An ordinate to the minor axis of an ellipse is to the 
corresponding ordinate of the inscribed circle, as the major 
axis is to the minor axis. 

Let a circle be described on BB' as a di- 
ameter, and let the ordinate PR of the el- 
lipse meet the circumference of the circle 
in P'. 

The equation of the ellipse when the ori- J 
gin is at the centre is 

But (b—y)(b+y) represents BRxB'R; hence Ave have 




BE 2 



a 



BRxB'B-F 

ButBRxB'R=P'R 2 ; hence 

PR 2 a" 



or 



P'R'-J" 
PR:P'R::a:3::2a:2J. 



134. To find the equation to the tangent at any point of an 
ellipse. 
Let the equation to the ellipse he a a y*+JV=a'5 a . 



112 



ANALYTICAL GEOMETRY. 



Let x\ y' be the co-ordinates of the point on the curve at 
which the tangent is drawn, and x", y" the co-ordinates of an 
adjacent point on the curve. The equation to the secant line 
passing through the points x\ y' and #", y" (Art. 40) is 

'y-y'=P^,(x-x'). (1) 

Now, since the points x', y' and x", y" are both on the ellipse, 

we must have a'y n +JV 2 = a 2 5 a , 

and fly"+JY"=«V; 

therefore, by subtraction, a > {7/" i -y n ) + b\x m -x' i )-0, 

y"-y' tf x"+x' 
or // / — —" o • // • /• 

x —x a y +y 

Substituting this value in equation (1), the equation of the 
secant line becomes 

y -^ s= "l'y w +F (<B " a0 (2) 

The secant will become a tangent when the two points coin- 
cide, in which case 

x r — x f \ and y f —y'\ 
Equation (2) will then become 

Vx' 

y-y r --^ x ~ x ')> ( 3 ) 

which is the equation to a tangent at the point x\ y\ 

Clearing this equation of fractions and transposing, we ob- 
tain cfyy' + tfxx^ay^ + tfx" ; 
hence a' i yy' + b' i xx' = a' 2 b% (4) 
which is the simplest form of the equation to the tangent line. 

135. Points where the tangent cats the axes. In equation 

(4) of the last article, x and y are 
co-ordinates of any point of the tan- 
gent line. Make y= 0, in which 
case # = CT, and we have 
If XX' ' — a^Tf '; 




that is, 



~x'' 



THE ELLIPSE. 113 

But x' is CE ; hence CR . CT = C A 2 . 

If from CT we subtract Cli or a?', we shall have the subtangent 

a* , cf-x'* 



~KY = -,-x 



X x 

Since the subtangent is independent of the minor axis, it is the 
same for all ellipses which have the same major axis ; and 
since the circle on the major axis may be considered as one of 
these ellipses, the subtangent is the same for an ellipse and its 
circumscribing circle. 

To determine the point in which the tangent intersects the 
axis of Y, we make x=0, which gives 

y 

Therefore CN.CT'^CB 2 . 

136. To draw a tangent to an ellipse through a given point. 
Let P be the given point on 

the ellipse. On AA ; describe 
a circle, and through P draw 
the ordinate PR, and produce 
it to meet the circumference of 
the circle in P'. Through P' 
draw the tangent P ; T, and from 
T, where the tangent to the cir- 
cle meets the major axis pro- 
duced, draw PT ; it will be a tangent to the ellipse at P (Art. 
135). 

137. To find the equation of a tangent line to the ellipse in 
terms of the tangent of the angle it makes with the major axis. 

In the equation of the tangent line (Art. 134, Eq. 3), 

Ux' 
--y represents the trigonometrical tangent of the angle 

which the tangent line makes with the major axis of the el- 




114 ANALYTICAL GEOMETEY. 

lipse (Art. 40). If we represent tins tangent by m, we shall 
have 

Vx' 

The equation of the tangent line (Art. 134, Eq. 4) was re- 
duced to the form 

a*yy' +&<$%' =a*b\ 

Vxx' V 
Hence y——— ir - r J r — 

ay y 

b* 

or y—mx+—. 

We wish then to express — in terms of m. 

Now # V = — cfy'm, 

and ay 2 + Vx"=a?b 2 ; 



j**, 



Therefore aY+^^-^aV. 

Hence y'\a*m*+V)=b\ 

and -p = ± Vcfm* + V\ 

Hence the equation to the tangent may be written 

y = mx ± 'y/cfm* + b 2 . 
Hence the straight line whose equation is 
y = mx ± Vcc 2 m 2 + b% 
touches the ellipse whose equation is a?y 2 +b' 2 x 2 — a?b' i . 

Since m in this equation is indeterminate, it may assume 
successively any number of values. The corresponding straight 
lines w T ill be a series of tangents to the ellipse. The double 
sign of the radical shows, moreover, that for any value of m 
there are two tangents to the ellipse parallel to each other. 

Ex. 1. In an ellipse whose major axis is 50 inches, the ab- 
scissa of a certain point is 15 inches, and the ordinate 1G inch- 
es, the origin being at the centre. Determine where the tan- 
gent passing through this point meets the two axes produced. 

Ans. Distance from the centre on the axis of X, = 41§ inch- 
es ; on the axis of Y, =25 inches. 



THE ELLIPSE. 



115 



Ex. 2. Find the angle which the tangent line in the preced- 
ing example makes with the axis of X. Am. 30° 57'. 

Ex. 3. On an ellipse whose two axes are 50 and 40 inches, 
find the point from which a tangent line must be drawn in or- 
der that it may make an angle of 35° with the axis of X. 

Ex. 4. Find the equations of the two lines which touch the 
ellipse 25y 2 + lG t # 2 = 400, and which make an angle of 135° with 
the axis of X. j LnSt y — _#± VS. 

138. To find the equation to the normal at any point of an 
ellipse. 

The equation to a straight line 
passing through the point P, whose 
co-ordinates are x\ y' (Art. 38), is 
y-y' = m(x-x'); (1) 
and, since the normal is perpendic- 
ular to the tangent ; we shall have 
(Art. 45) ! 

m — t. 

— ml 

But we have found for the tangent line, Art. 137, 




5V 



m —- 



Hence 



on — 



ay 

ay 



Vx" 



Substituting this value in equation (1), we shall have for the 
equation of the normal line 

ah/ 

(2) 



y-y 



' lfx^ X ~~ X ^ 



where x and y are the general co-ordinates of the normal line, 
and x ', y' the co-ordinates of the point of intersection with the 
ellipse. 

139. Points of intersection with the axes. To find the point 
in which the normal cuts the major axis, make y—0 in equa- 
tion (2), and we have, after reduction, 

GN, or x— — x . 



116 



ANALYTICAL GEOMETRY. 



If we subtract .this value from CE, which is represented by a?', 
we shall have the subnormal 



a 






To find the point in which the normal cuts the minor axis, 
make x—0 in equation (2), and we have 

140. Distance from the focus to the foot of the normal. If 
a?-U 



w r e put e* for 



a 



(Art. 127), we shall have 
GN"=sV. 



If to this we add F'C, which equals c or ae (Art. 127), we have 

F'N" = ae + e*x' = £^ + *#')> 
which is the distance from the focus to the foot of the normal. 

Ex. 1. In an ellipse whose major axis is 50 inches, the ab- 
scissa of a certain point is 15 inches, and the ordinate 16 inch- 
es, the origin being at the centre. Determine w T here the nor- 
mal line passing through this point meets the two axes. 

An$. Distance from the centre on the axis of X, = 5-J inch- 
es ; on the axis of Y, = 9 inches. 

Ex. 2. Find the point on the curve of an ellipse whose two 
axes are* 50 and 40 inches, from which, if an ordinate and nor- 
mal be drawn, they will form with the major axis a triangle 
whose area is 80 inches. 



141. The normal at any point of an ellipse bisects the an- 
gle formed by lines drawn from that point to the foci. 

Let PT be a tangent to an el- 
lipse, and PF, PF' be lines drawn 
from the point of contact to the 
foci. Draw PJST bisecting the an- 
gle FPF'. Then, by Geom., Bk. 
IV., Pr. 17, 

FP:FP::EN:FNj 
or by composition, 




THE ELLIPSE. 



117 



FP + FT : FF' : : FT : F'N. (1) 

But FP+FT = 2a. 

Also FF / = 2c=2ae\Avt 127), 

and FT=a+ex (Art. 128). 

Making these substitutions in proportion (1), we have 
2a:2ae::a+ex: F'N. 
Hence F'JST = e(a + ex). 

But by Art. 140, e(a + ex) represents the distance from the fo- 
cus F' to the foot of the normal. Hence the line PN ; which 
bisects the angle FPF' ? is the normal. 

142. The radii vectores are equally inclined to the tangent. 
Since PN is perpendicular to TT', and the angle FPN is equal 
to the angle FTN, therefore the angle FPT is equal to the an- 
gle ftt'. 



143. Second method of drawing a tangent line to an ellipse. 

Let P be the point through 
which the tangent line is to be 
drawn. Draw the radii yectores 
PF, PF / ; produce PF' to G, 
making PG equal to PF, and 
draw FG-. Draw PT perpendic- 
ular to FGr, and it will be the 
tangent required; for the angle 
FPT equals the angle GPT, which equals the vertical angle 
FTT'. 




144. Every diameter of an ellipse is bisected at the centre. 

Let PP' be a straight line drawn 
through the centre of the ellipse, 
and terminated on both sides by the 
curve ; it will be divided into two 
equal parts at the point C. Let x\ 
y' be the co-ordinates of the point P, 
and x" , y" those of the point P'. 




118 



ANALYTICAL GEOMETRY. 



Since the points P and P' are on the curve, we shall have 
(Art. 121) 



and 



y">~(a<-x'»); 



y' 2 a?—x' 2 
whence, by division, —j^——. m . 

But, since the right-angled triangles CPR, CP'R' are similar, 
we have 






x 



x ' 



IIcnce a^-aW" 

Clearing of fractions, we obtain 

^'2/72. 
tO i/O , 

whence also we have y^—y"*. 

Consequently, x" 1 + y n = x" 2 + y f 

or CP 2 = CP /2 ; 

that is, CP=CF; 

that is, PP / is bisected in C. 



145. Tangents to an ellipse at the extremities of a diameter 
are parallel to each other. 

a 2 




In Art. 135 we found CT: 



a 



x 



and similarly CT' = — , where x' 



x 



represents CR, the abscissa of 
the point P, and x" represents 
CR/, the abscissa of the point P'. But we have found (Art. 
144) that x' = x n ; hence CT = CT'. The two triangles CPT, 
CP'T', have therefore two sides, and the included angle of the 
one equal to two sides and the included angle of the other; 
hence the angle CPT = the angle CP'T', and PT is parallel to 

p/rjv 




THE ELLIPSE. 119 

Hence, if tangents are drawn through the vertices of any 
two diameters, they will form a parallelogram circumscribing 
the ellipse. 

146. If from any point in the curve, chords are drawn to 
the extremities of the major axis, the product of the tangents 
of the angles which they form icith it, on the same side, is 

equal to — — . 

Let PA, PA 7 be two chords 
drawn from the same point, P, 
on the ellipse to the extremities 
of the major axis. 

The equation of the line PA, 
passing through the point A, 
whose co-ordinates are x' ~a, 
t/=0 (Art. 38), is 

y=?7i(x—ct). 

The equation of PA 7 , passing through the point A 7 , whose 
co-ordinates are x" ^ — a, y n — 0, is 

y=zm'(x+a). 

At the point of intersection, P, these equations are simulta- 
neous, and, combining them together, we have 

y 1 z=z mm'(x 2 — a r ). (1) 

But, since the point P is on the curve, we must have at the 
same time 

y'—tf-x^-^-a*). (2) 

Comparing equations (1) and (2), we see that 

, v 

mm ■= — — , 
a'' 

where m denotes the tangent of the angle PAX, and m' de- 
notes the tangent of the angle PA'X. 

147. Definition. Two chords drawn from any point in the 
curve to the extremities of a diameter are called supplementa- 
ry chords. 




120 ANALYTICAL GEOMETRY. 

148. Supplementary chords in a circle. A circle may be 
considered as an ellipse whose two axes are equal to each oth- 
er ; hence, in a circle, 

mm'= — 1, 
which shows that the supplementary chords are perpendicular 
to each other (Art. 46). 

149. If through one extremity of the major axis a chord be 
drawn parallel to a tangent line to the curve, the supplement- 
ary chord will be parallel to the diameter drawn through the 
point of contact, and conversely. 

Let DT be a tangent to the 
ellipse at the point D, and let 
the chord AP be drawn parallel 
to it ; then will the supplement- 
ary chord AT be parallel to 
the diameter DD', which passes 
through the point of contact, D. 
Let a?', y f designate the co-ordinates of D. The equation of 
the line CD (Art. 31) gives 

y r 

whence m r — —,. 

x 

But the tangent of the angle which the tangent line makes 

with the major axis (Art. 137) is 

JV 

ay 
Multiplying together the values of m and m', we obtain 

, v 

mm = 5, 

€T 

which represents the product of the tangents of the angles 
which the lines CD and DT make with the major axis pro- 
duced. 

But, by Art. 14G, the product of the tangents of the angles 

V 
PAT and PAT is also equal to — —. Hence, if AP is paral- 
lel to DT, AT will be parallel to CD, and conversely. 




THE ELLITSE. 121 

150. Definition. Two diameters of an ellipse are said to be 
conjugate to one another when each is parallel to a tangent 
line drawn through the vertex of the other. 

151. Property of conjugate diameters. Let DD' be any di- 
ameter of an ellipse, and 
DT the tangent line drawn 
through its vertex, D, and 
let the chord AP be drawn 
parallel to DT; then, by 
Art. 1-19, the supplementa- 
ry chord AT is parallel to 
DD'. Let another tangent, ET', be drawn parallel to AT ; it 
will also be parallel to DD'. Let the diameter EE' be drawn 
through the point of contact, E ; then, by Art. 149, AT being 
parallel to T'E, the supplementary chord AP, and also its par- 
allel DT, will be parallel to EE'. Hence each of the diameters 
DD', EE' is parallel to a tangent drawn through the vertex 
of the other, and by definition (Art. 150) they are conjugate to 
one another. 

Since the conjugate diameters DD', EE' are parallel to the 
supplementary chords AT, AP, by Art. 146, the product of the 
tangents of the angles which conjugate diameters form with 

V 
the major axis is equal to — — . 

Ex. 1. In an ellipse whose axes are 10 and 8, a chord drawn 
from one extremity of the major axis forms with that axis an 
angle whose tangent is 2 ; what angle does the supplementary 
chord form ? Arts. 

Ex. 2. In an ellipse whose axes are 12 and 8, a chord forms 
with the major axis an angle whose tangent is — 3 ; what angle 
does the supplementary chord form ? Ans. 

Ex. 3. In an ellipse whose axes are 10 and 8, find the angles 
which supplementary chords drawn from the point x—1 form 
with the major axis. Ans. 

Ex. 4. In an ellipse whose axes are 10 and 30, two conjugate 

F 



122 ANALYTICAL GEOMETRY. 

diameters are equally inclined to the major axis. Find the 
angle between the two diameters. 

152. To determine the co-ordinates of the joints of intersec- 
tion of a straight line with an ellipse. 

Let the equation to the ellipse be 

ay + b 2 x*=a*b% (1) 

and the equation to a straight line be 

y=mx + c. (2) 

If this line intersects the ellipse, then we may regard (1) and 
(2) as simultaneous equations containing but two unknown 
quantities. ' By substitution in equation (1) we obtain 

(> 2 m 2 + 5> 2 + 2a 2 c?nx = (b 2 - c 2 )a% 
the roots of w T hich equation give the abscissas of the points 
w r here the straight line meets the curve, and the ordinates may 
be found from equation (2). Hence, if the roots be real, the 
straight line will cut the ellipse in two points, and it can not 
cut the ellipse in more than tw^o points. If the roots are equal, 
the points of section coincide, and the line is then a tangent. 
If the roots are imaginary, the line falls entirely without the 
ellipse. 

Ex. 1. Find the co-ordinates of the 
points in which the ellipse whose equa- 
tion is 25y 2 + 16a? 2 =:400 is intersected 
by the line whose equation is y— 2x— 5. 
Ans. #=+3.7999, or +0.5104; 
y= +2.5998, or -3.9792. 
Ex. 2. Find the co-ordinates of the 
points in w T hich the ellipse w T hose equa- 
tion is £9?/ 2 — 36# 2 =zl764 is intersected by the line w r hose equa- 
tion is y=z3x—7, and draw a figure representing the several 
quantities. 

153. To determine the co-ordinates of the points of inter- 
section of a circle and ellipse. 

If the centre of the circle is not restricted in position, there 




THE ELLIFSE. 123 

may be four points of intersection corresponding to an equa- 
tion of the fourth degree. If, however, the centre of the cir- 
cle is at one extremity of the major axis, there will be but two 
points of intersection, which may be easily found. 
Let the equation to the ellipse be 

V 

y 2 =-£lax—x 2 \ 

CO 

and the equation to the circle be 

x* + y 2 =r 2 ; 
then, by substitution, we obtain 

V 

r 2 —x 2 = —(2ax — x 2 ), 

where x will be found to have tw T o values, but one of them is 
negative, and gives imaginary vqjues for y. There will, there- 
fore, be but two points of intersection, both having the same 
abscissa, and the ordinates w T ill differ only in sign. 

Ex. 1. Find the co-ordinates of the points in which the el- 
lipse whose equation is y 2 —^(10x— x 2 ) is intersected by the 
circle whose equation is x 2 -\-y 2 — §k. 

Ex. 2. Find the co-ordinates of the points in w r hich the el- 
lipse wliose equation is y 2 = ^(14:X— x 2 ) is intersected by the 
circle w T hose equation is x 2 + y 2 =zl00. 

If the centre of the circle is upon either axis of the ellipse, 
there may be four points of intersection. 

Ex. 3. Find the co-ordinates of the points where the ellipse 
y 2 — ^_(100— x 2 ) is intersected by the circles x 2 +y 2 = 64:; y 2 + 
(#_2) 2 = 64, y * + (x-8y = 64:, and ?/ 2 + (£-20) 2 = 64. 

The first circle cuts the ellipse in four points, the second cuts 
it in three points, the third in two points, and the fourth does 
not cut the ellipse. 

Ex. 4. Draw a figure representing these curves and their in- 
tersections. 

154. Having given the co-ordinates of one extremity of a 
diameter, to find those of either extremity of the diameter con- 
jugate to it. 



124 



ANALYTICAL GEOMETRY. 



fl 

1 




PK 


I) 


\ N 






] 


^\__ 




>i' 



Let AA 7 , BB 7 be the axes of an el- 
lipse; DD 7 , EE 7 a pair of conjugate 
diameters. Let x', y' be the co-ordi- 
A nates of D ; then the equation to CD 
(Art. 40) is y ' 



* x' 



(1) 



Since the conjugate diameter EE' is parallel to the tangent at 
D, the equation to EE' (Art. 149) is 

Vx' 



y=- 



ay 



7 • X» 



(2) 



To determine the co-ordinates of E and E 7 , we must combine 
the equation to EE 7 with the equation to the ellipse, cfy* +5V 
=tfV. 
Substituting the value of y&om equation (2), we have 



Therefore 
or 

whence 
and 



ay 



x' = 



ay* 



x=± 



ay' 



Taking the minus sign, in which case x is CN, and combining 
with equation (2), we have 



a 



We thus find the co-ordinates of the point E. The co-ordinates 
of the point E' have the same values with .contrary signs. 

155. The sum of the squares of any two conjugate diame- 
ters is equal to the sum of the squares of the axes. 
Let x', y' be the co-ordinates of D ; then, by Art. 154, 

«y . Vx n 






CD 5 + CF 



:ar"+y"4 



+• 



V a; 



a 



=a"^-h\ 






THE ELLIPSE. 



125 



156. The rectangle contained by the focal distances of any 
point on the ellipse is equal to the square of half the corre- 
sponding conjugate diameter. 

Let DD', EE' be a pair of conjugate 
diameters, and from D draw lines to the 
foci, F and F'. Represent the co-ordi- 
nates of D referred to rectangular axes 
by x\ y. 

Then, since CD 2 + CE 2 =a 2 + 6 2 (Art. 
155), we have 

CE 2 =a 2 +6 2 -CD', 




=a>-(l-^>>, 



=^ 2 -^V 2 (Art.l27), 
=DFxDF'(Art.l28); 

that is, the product of the focal distances DF, DF' is equal to 
the square of half EE', which is the diameter conjugate to the 
diameter which passes through the point D. 



157. The parallelogram formed by drawing tangents through 
the vertices of two conjugate diameters is equal to the rectan- 
gle of the axes. 

Let DD',EE' be two con- 
jugate diameters, and let D 
ED'E' be a parallelogram 
formed by drawing tangents 
to the ellipse through the 
extremities of these diame- 
ters ; the area of the paral- 
lelogram is equal to AA' x BB'. 

Draw DM perpendicular to EE', and let the co-ordinates of 
D referred to rectangular axes be x', y '. 

The area of the parallelogram DED'E' is equal to 4CE . DM, 




126 ANALYTICAL GEOMETRY. 

which is equal to 4CE.CT sin. CTG, which is equal to 4CT. 
EN, because EC and DT are parallel. 

But CT=- (Art. 135), and EN=— (Art. 154) ; 

X CO 

a? bx f 
hence the parallelogram DED'E' = 4 . - . — = 4a6 = AA' x BB'. 

Ex. 1. In an ellipse whose axes are 10 and 8, what is the 
length of a diameter which makes an angle of 45° with the 
axis of x ? What is the length of its conjugate ? 

Ans. 

Ex. 2. What is the altitude of the circumscribed parallelo- 
gram whose sides are parallel to the conjugate diameters of the 
preceding example ? Ans. 

158. Equation to the ellipse referred to a pair of conjugate 
diameters as axes. 

Let CD, CE be two conjugate semi- 
diameters ; take CD as the new axis of 
x, CE as that of y; let DCA=a, EC A 
=/3. Let x, y be the co-ordinates of 
any point of the ellipse referred to the 
original axes, and x f , y' the co-ordinates 
of the same point referred to the new axes. 

The equation of the ellipse referred to its centre and axes 
(Art. 121) is ay + 5W = a%\ 

In order to pass from rectangular to oblique co-ordinates, the 
origin remaining the same, we must substitute for x and y in 
the equation of the curve (Art. 56) the values 
x—x f cos. a+y' cos. /3, 
y=x' sin. a + y' sin. /3. 
Squaring these values of *x and ?/, and substituting in the equa- 
tion of the ellipse, we have 

x'\a* sin. 2 a + V cos. 2 a) + y'\a* sin. 2 /3 + V cos. 2 /3) + 
2x'y'(a' 2 sin. a sin. fi + b 2 cos. a cos. /3) = a 2 i 2 , 
which is the equation of the ellipse when the oblique co-ordi- 
nates make any angles a, j3 with the major axis. 




THE ELLIPSE. 127 

But, since CD, CE arc conjugate semidiameters, we must 
have (Art. 151) 

b 2 
mm'= tang, a tang. ]3= — — , 

whence a 2 tang, a tang. /3 + b 2 = 0. 

Multiplying by cos. a cos. /3, remembering that cos. a tang, a 

= sin. a, we have 

a 2 sin. a sin. j3 + & 2 cos. a cos. fi = 0. 
Hence the term containing x'y' vanishes, and the equation be- 
comes 

x'Xc? sin. 2 a + V cos. 2 a) + y f \a 2 sin. 2 j3 + 6 2 cos. 2 j3) = a*b% (1) 
which is the equation of the ellipse referred to conjugate diam- 
eters. 

If in this equation we suppose y r — 0, we shall have 

a*b 2 

tAs 2 * 2 ■ 7 2 Q • 

a Sin. a + o cos. a 
This is the value of CD 2 , which we shall denote by a' 2 . 
If we suppose x' = 0, we shall have 

y ~^ 2 sin. 2 /3 + 6 2 cos. 2 /3- 
This is the value of CE 2 , which we shall denote by b /2 . 

Dividing equation (1) by a 2 b 2 , and then substituting for the 

coefficients of x' 2 and y' 2 the equal values -75 and -775, we have for 

the equation to the ellipse referred to conjugate diameters 

x f2 y t2 H 

— 4- — — 1 • 

or, suppressing the accents of the variables, we have 

a /2 + b /2 

159. 7%^ square of any diameter is to the square of its con- 
jugate, as the rectangle of the parts into which it is divided by 
any ordinate, is to the square of that ordinate. 

The equation of the ellipse referred to conjugate diameters 
may be put under the form 

a f2 y 2 = V\a' 2 -x 2 ). 



128 



ANALYTICAL GEOMETKY. 




This equation may be reduced to the 
D proportion 

a n \b'*\\a n -x"\y\ 
or (2a') 9 : {2bJ : : (a'+x)(a'-x) : y\ 
Now 2a' and 2V represent the conju- 
> e / gate diameters DD', EE' ; and, since x 
represents CR, a'+x will represent D'R, and &'— x will repre- 
sent DR; also PR represents y; hence 

DD' 2 : EE /a : : DR x RD' : PR 2 . 
If we draw a second ordinate P'R' to the diameter DD', we 
shall have 

PR 2 : DR x RD' : : V* :a":: POT : DR' x R'D', 
or PR 2 : P'R' 2 : : DR x RD 7 : DR' x R'D' ; 

that is, the squares of any two ordinates to the same diameter 
are as the products of the parts into which they divide that 
diameter. 

160. To find the polar equation to the ellipse, the pole being 
at one of the foci. 

1. Let F be the pole. 
Let FP=r; angle PFA=0; then 
FR=t> cos. 9. 

By Art.128, r = a-ex. 

tf=:CR=:CF + FR, 

=ae+r cos. 0. 

Therefore r=a—a£—er cos. 0. 

r(l + ecos.O) = a(l-e 2 ), 

a(l-e*) 
fp — — - — 

1 + e cos. 0' 

which is the required equation when 6 is measured from the 
radius to the nearer vertex. 
2. Let F' be the pole. 

Let FT=// PF'A=0'; then F / E=r / cos. 0'. 
By Art.128, r'=a+ex. 

But &=CR=E'K-F'C, 

— r' cos. O'—ae. 







R/ A But 






THE ELLIPSE. 129 

Therefore r' = a+er' cos. O'—ae 2 . 

Hence r\l—e cos. 0') = a(l — e 2 ) y 

or ^-i-^cos.r 

which is the required equation when 0' is measured from the 
radius to the remote vertex. 

Ex. 1. The axes of an ellipse are 50 and 40 inches, and the 
radius vector is 12 inches. Determine the value of 0. 

Ans. 56° 15'. 

Ex. 2. The axes of an ellipse are 50 and 40 inches, and is 
equal to 36°. Determine the radius vector. 

• Ans. 10.771 inches. 

Ex. 3. In an ellipse whose major axis is 50 inches, the radius 
vector is 12 inches, and 9 is 36°. Determine the minor axis 
of the ellipse. Ans. 41.67 inches. 

161. Any chord which passes through the focus of an ellipse 
is a third proportional to the major axis and the diameter 
parallel to that chord. 

Let PP ; be a chord of an ellipse 
passing through the focus F, and let / / N^p 

DD' be a diameter parallel to PP'. J 

By Art. 160, TF=r=^P-^- h . 
J ' 1 + 6 cos. 

To find the value of FP', we must 

substitute for 0, 180° + 0, and we obtain 

0(1-6*) 




FP' = *>'=: 



1—e cos. 6' 



Hence ?F'=r+r' = , 2ai l \ . (1) 

1—e cos. v ' 



But, by Art. 158, „«, 

CD" 



a 2 sin.*d + b* cos.'d* 

a a sin. 2 0+(a 2 -aV)cos. a (Art 127) ' 
a'b* 



V-aVcos.*0' 
F2 



(2) 



130 ANALYTICAL GEOMETRY. 

a\l-e*) 
~l-£ 2 cos. 2 0' 
Comparing equations (1) and (2), we find 
_> 2CD 2 4CD 2 

pp — =-o — ; 

a 2a ' 

that is, AA':DD'::DD':PP', 

or PP' is a third proportional to AA' and DD'. 

162. Definition. The parameter of any diameter is a third 
proportional to that diameter and its conjugate. 

The parameter of the major axis is called the principal pa- 

2V 
rameter, or latus rectum, and its value is — (Art. 126). The 

2a 2 a 

parameter of the minor axis is -y-. The latus rectum is the 

double ordinate to the major axis passing through the focus 
(Art. 126). Now, since any focal chord is a third proportional 
to the major axis and the diameter parallel to that chord, and 
since the major axis is greater than any other diameter, it fol- 
lows that the major axis is the only diameter whose parameter 
is equal to the double ordinate passing through the focus. 

163. Definition. The directrix of an ellipse is a straight 
line perpendicular to the major axis produced, and intersecting 
it in the same point with the tangent drawn through one ex- 
tremity of the latus rectum. 

Thus, if LT be a tangent drawn through one extremity of 
the latus rectum LI/, meeting the major axis produced in T, 
and NT be drawn through the point of intersection perpendic- 
ular to the axis, it will be the directrix of the ellipse. 

The ellipse has two directrices, one corresponding to the fo- 
cus F, and the other to the focus F'. 

164. The distance of any point in an ellipse from either fo- 
cus is to its distance from the corresponding directrix, as the 
eccentricity is to unity. 






THE ELLIPSE. 



131 




Let F be one focus of an 
ellipse, NT the corresponding 
directrix ; F' the other focus, 
and N'T' the corresponding 
directrix. Let P be any point 
on the ellipse ; x, y its co-or- 
dinates, the centre being the origin. Join PF, PF', and draw 
NPN' parallel to the major axis, and PB perpendicular to it. 

By Art. 135, CT=-=-. 

Hence, subtracting CB or x, 

e e 

But, by Art. 128, r=FF=a-ex. 
Hence *.BT, or <?.PN=PF; 

or, PF:PN::<?:1. 

In like manner, we find that 

PF':PN'::*:i. 




165. To find the area of an ellipse. 

On A A', the major axis of an el- 
lipse, let a semicircle be described, 
and within this semicircle inscribe a 
polygon, AMM'A'. From the ver- 
tices of this polygon draw ordinates 
to the major axis, and join the points in which they intersect 
the ellipse, thus forming a polygon ANN 'A', having the same 
number of sides. 

Let Y, Y', etc., denote the ordinates of the points M, M', etc., 

and let y, y ', etc., denote the ordinates of the points N, N', etc., 

corresponding to the same abscissas x, x\ etc. 

Y+Y' 
The area of the trapezoid RMIFE^ — ^ — (x— x'), 

y-\-y r 
and the area of the trapezoid KN^B/— ' ' (x— x f ). 

Hence EMM'B' : ENN'E' : : Y + Y' : y+y f . 



132 ANALYTICAL GEOMETRY. 

But, by Art. 132, Y:y::a:b; 
also Y' : y' : : a : b. 

Whence Y + Y' :y+y' : :a: b. 

Therefore RMM'R 7 : RNN'R' : : a : b. 

In the same manner it may be proved that each of the trap- 
ezoids composing the polygon inscribed in the circle, is to the 
corresponding trapezoid of the polygon inscribed in the ellipse, 
as a is to b ; hence the entire polygon inscribed in the circle is 
to the polygon inscribed in the ellipse, as a is to b ; and this 
will be true whatever be the number of sides of the polygons. 

If now the number of sides be indefinitely increased, the 
areas of the polygons will become equal to the areas of the 
semicircle and semi-ellipse respectively, and we shall have the 
first is to the second as a is to b ; or, denoting the area of the 
circle by S, and that of the ellipse by s y we shall have 

S : s : : a : b ; that is, s= -S. 

But the area of a circle whose radius is a is represented by 
7iV; hence • s — irab; 

or the area of an ellipse is equal to n times the rectangle de- 
scribed upon its semi-axes. 



166. Since irab— -vArV^^: Vira* x 7rb% we find that the area 
of an ellipse is a mean proportional between the areas of its 
circumscribed and inscribed circles. 

Ex. 1. Determine the area of an ellipse whose two axes are 
24 and 18 inches. 

Ex. 2. The area of an ellipse is 40 square inches, and the la- 
tus rectum is 4 inches ; required the axes of the ellipse. 

Ex. 3. The axes of an ellipse are 40 and 50 ; find the areas 
of the two parts into which it is divided by the latus rectum. 



THE HYPERBOLA. 



133 



SECTION VII. 



THE HYPERBOLA. 




167. An hyperbola is a plane curve traced out by a point 
which moves in such a manner that the difference of its dis- 
tances from two fixed points is always the same. The two 
fixed points are called the foci of the hyperbola. 

Thus, if F and F' are two fixed points, and if the point P 
moves about F in such a manner that 
the difference of its distances from F 
and F' is always the same, the point P 
will describe an hyperbola, of which 
F and F' are the foci. 

If the point P' moves about F' in 
such a manner that P'F— P'F 7 is al- 
ways equal to PF'— PF, the point P' 
w T ill describe a second portion of the curve similar to the first. 
The two portions are called branches of the hyperbola. 

The distance of the point P from either focus is called the 
focal distance, or the radius vector. 

168. Mechanical description of the curve. From the defi- 
nition of an hyperbola the curve may be described mechanic- 
ally. Take any two points, as F and 
F'. Take a ruler longer than the 
distance FF', and fix one of its ex- 
tremities at the point F 7 so that the 
ruler may be turned round this point 
in the plane of the paper. Take a 
thread shorter than the ruler, and 
fasten one end of it at F, and the 
other to the end M of the ruler. Then move the ruler on its 




134 ANALYTICAL GEOMETRY. 

pivot at F' 3 while the thread is kept constantly stretched by a 
pencil pressed against the ruler ,• the curve described by the 
point of the pencil will be a portion of an hyperbola. For in 
every position of the ruler, the difference of the distances from 
the variable point P to the two fixed points F and F' will al- 
ways be the same, viz., the difference between the length of the 
ruler and the length of the thread. 

If the ruler be turned and move on the other side of the 
point F, the other part of the same branch may be described. 

Also, if one end of the ruler be fixed at F, and that of the 
thread at F', the opposite branch of the hyperbola may be de- 
scribed. 

169. The centre of the hyperbola is the middle point of the 
straight line joining the foci. 

A diameter is any straight line passing through the centre, 
and terminated on both sides by opposite branches of an hy- 
perbola. 

The diameter which, when produced, passes through the foci, 
is called the transverse axis. 

The latus rectum is the chord drawn through one of the foci 
perpendicular to the transverse axis. 

170. To find the equation to the hyperbola. 

Let F and F ; be the foci, and draw the rectangular axes CX, 

CY, the origin C being placed at the 
middle of FF'. Let P be any point 
of the curve, and draw PR perpen- 
dicular to CX. 

Let 2c denote FF ; , the constant 
distance between the foci, and let 2a 
denote FT— FP, the constant dif- 
ference of the focal distances. De- 
note PF by r, PF' by r' } and let x and y denote the co-ordinates 
of the point P. 

Then, since FP 2 = PR 2 + RF = PR 2 + (CR - CF) a , 




THE HYPERBOLA. 135 

we have 7'*=y* + (x— c)\ (1) 

Also, PF /2 = PE 2 + EF /2 = PE 2 + (CE + CF) 2 ; 

that is, r'*=tf+(x+c)*. (2) 

Adding equations (1) and (2), we obtain 

r* + r' 2 = 2(?/* + x* + c>); (3) 

and subtracting equation (1) from (2), we obtain 

which may be put under the form 

(r' + r)(r'—r) — ±cx. (4) 

But, from the definition of the hyperbola, we have 

r'-= r—2a. 
Substituting this value in equation (4), we obtain 

2cx 
a 
Combining the last two equations, we find 

r=-a+-. (6) 

Squaring these values, and substituting them in equation (3), 
we obtain 

which may be reduced to the form 

(c 3 -a> 2 -^y=:a 2 (c 2 -a 2 ), (7) 

which is the equation to the hyperbola. 
If we put V — &^- # 2 , the equation becomes 

Vx*-ay = c?b\ (8) 

x 2 if 

tf-ih 1 ' (°) 

which is the equation to the hyperbola referred to its centre 
and transverse axis. 

This equation is sometimes written 

pJ^rltf). (io) 

The equation to the ellipse becomes the equation to the hy- 



136 ANALYTICAL GEOMETRY. 

perbola.by writing — & a for 5 2 ; and we shall find that the hy- 
perbola has many properties similar to those of the ellipse. 

171. Points of intersection with the axes. To determine 
where the curve cuts the axis of X, make y=0 in the equation 
to the hyperbola, and we obtain 

x=dza=:GAor CA', 
which shows that th$ curve cuts the axis of abscissas in two 
points, A and A 7 , at the same distance from the origin, the one 
being on the right, and the other on the left ; and, since 2CA, 
or AA',is equal to'2a, it follows # that the difference of the two 
lines drawn from any point of an hyperbola to the foci, is 
equal to the transverse axis. 

If we make x=0 in the equation of the hyperbola, we ob- 
tain 

which shows that the hyperbola does not intersect the axis CY. 

172. If with A or A' as a centre, and a radius equal to CF, 
we describe a circle cutting the axis of y in two points, B and 
B', we shall have 

CB 2 =BA 2 -CA a 

that is, £=CBorCB'. 

The line BB' thus determined is called the conjugate axis of 

the hyperbola. 

173. Figure of the hyperbola determined. In equation (10), 
Art. 170, let x be numerically less than a ; then the values of y 
are imaginary; therefore no point of the hyperbola is nearer 
the axis of y than ±.a. 

Let x be numerically greater than a; then for each value of 
x there are two equal values of y with contrary signs. 

As x increases, the values of y increase $ and when x be- 
comes indefinitely great, the value of y becomes so likewise. 

The hyperbola therefore consists of two opposite branches, 



THE HYPERBOLA. 



137 



extending indefinitely to the right of A and to the left of A', 
and symmetrically placed with respect to the axis XCX'. 

174. Other points of the curve determined. If we wish to 
determine other points of the curve, we reduce the equation 
to the form 

from which we may compute the value of y corresponding to 
any assumed value of x. 

Ex. Trace the curve whose equation is 36%*— 49?/' = 1764. 

Solving the equation for y, we have 

6 ,-* 

2/=±^Vtf-49. 

If x be assumed less than 7, the corresponding value of y is 
imaginary. If we assume for x different values from 7 up- 
ward, we obtain the corresponding values of y as given below. 

When 0=11, y=± 7.27. 



0=12, y=± 8.35. 
x=13,y=±; 9.39. 
= 14, 2/= ±10.39. 



When 0= 7, y=0. 

0= 8,y= ±3.32. 
0= 9, 2/= ±4.85. 
£=10,2/= ±6.12. 

When a?=7, y=0, which gives 
the point A. When 0=8, y=± 
3.32, which gives two points, a and 
a\ one above and the other below 
the axis of X; when 0=9, 2/= ± 
4.85, which gives the points b and 
V ; when 0=lO,y =±6.12, which 
gives the points c and c\ etc. 

If we ascribe to x a negative value, we shall obtain for y the 
same pair of values as when we ascribed to the correspond- 
ing positive value. Hence the portion of the curve to the left 
of the axis of Y is similar to the portion to the right of it. 
Moreover, there is no point of the curve between the values 
0=+7 and 0=— 7. 





138 



ANALYTICAL GEOMETRY. 



175. To find the foci of an hyperbola when the two axes are 
given. Since J 2 = c 2 — & 2 , we have 

c 2 orCF 2 =a 2 + 3 2 =AB 2 ; 
that is, the distance from the centre to 
either focus of an hyperbola is equal to 
the distance between the extremities of its 
axes. 

If, then, from the centre 0, with a ra- 
dius equal to the diagonal of the rect- 
angle upon the semiaxes, we describe an arc cutting the trans- 
verse axis produced in F and F', the two points of intersection 
will be the foci of the hyperbola. 




176. To find the length of the latus rectum. According to 
Art. 170, eq. 10, 



a 



Suppose x=c or CF; then 



tf=-tf-a>). 






But &—c^ — V, Art. 172 ; hence we have 

or a:b:b:y, 

and 2a:2b::2b:2y. 

But 2y here represents the double ordinate drawn through the 

focus, and is called the latus rectum, Art. 169 ; hence the latus 

rectum of any hyperbola is a third proportional to the trans- 

verse and conjugate axes. 



177. Equation to the hyperbola in terms of the eccentricity. 
The fraction - , which represents the ratio of CF to CA, or the 

distance from the centre to either focus divided by half the 
transverse axis, is called the eccentricity of the hyperbola. If 
we represent the eccentricity by e, then 



THE HYPERBOLA. 



139 



a 



But we have seen that 
hence 



or 



c* = a 7 + b 2 ; 



& 



Making this substitution, the equation of the hyperbola becomes 

which is the equation in terms of the eccentricity. 

178. To find the distance of any point on the curve from 
either focus. Equations (5) and (6) of Art. 170 are 

, ex 

a 1 

ex 

c a 

Substituting e for -, these equations become 

r' = ex+a, 
r=ex—a, 
which equations represent the distance of any point on an hy- 
perbola from either focus. 

Multiplying these values together, we obtain 
rr'=ze*of—a 9 3 
which is the value of the product of the focal distances. 

179. The conjugate hyperbola. Suppose an hyperbola to be 
described whose foci F and F' are at 
the same distance from the centre C 
as those of the curve hitherto de- 
scribed, but lie upon the axis CY in- 
stead of CX, and suppose the differ- 
ence of the distances of any point on 
the new curve from the two foci is 
2b instead of 2a; then, retaining the 
same axes of reference as before, we shall have for the new 
position of F and F', 

FP 2 =PB'+BF 3 =PB 2 +(CB-CF) a ; 




140 ANALYTICAL GEOMETRY. 

that is r 1 = x 2 + (y — c)\ 

Also, FT 2 = PR 2 + F'R 2 = PR 2 + (CR + CF) 2 ; 

that is, r n =x* + {y+c)\ 

Proceeding as in Art. 170, we find 

(<f - V) y *- 1 V = b\& - V). 
Putting a? for c 2 — £ 2 , the equation becomes 

b* 
or tf = tf(?t + <?), 

which is the equation to the new hyperbola. 

In this equation, suppose a?=0, and we have y—±.b; that is, 
the curve passes through the points B and B', and BB' is the 
transverse axis of the new curve. 

Suppose y=0, and we have x= ±a>/—^ 
which shows that the curve does not meet the axis of X, and 
AA' is the conjugate axis of the new curve (Arts. 171 and 172). 

This new hyperbola is called the hyperbola conjugate to the 
former. One hyperbola is therefore said to be conjugate to an- 
other, when the transverse and conjugate axes of the one hyper- 
tola are the conjugate and transverse axes of the other hyperbola. 

If y*J^-a>) 

be the equation of any hyperbola, then 

is the equation to the hyperbola conjugate to the former. The 
latter equation may be deduced from the former by writing 
— a 2 for a% and — £ a for b\ 

Ex. 1. Trace the curve whose equation is 3# 2 — 5y 2 = 15. 

Ex. 2. In a given hyperbola half the difference of the focal 
distances is 7, and half the distance between the foci is 9; what 
is the equation to the hyperbola? 

Ex. 3. What is the eccentricity of the hyperbola whose equa- 
tion is 9£ 2 -10?/ 2 :=144:? 

Ex. 4. What is the equation of an hyperbola whose conjugate 
axis is 6 and the eccentricity 1^ ? 



THE HYPERBOLA. 



141 




180. To find the equation of the hyperbola when the origin 
is at the vertex of the transverse axis. 
The equation of the hyperbola when 
the origin is at the centre is 

tf—tf-a*). (1) 

If the origin is placed at A, the 
ordinates will have the same value 
as when the origin was at the centre, 
but the abscissas will be changed. 

If we represent the abscissas reckoned from A by x\ we 
shall have CR = AR + AC, 

or x—x'-\-a. 

Substituting this value of x in equation (1), we have 

which is the equation of the hyperbola referred to the vertex A. 

181. Relation of ordinates to the transverse axis. If the 
last equation be resolved into a proportion, we shall have 

y':(2a+x)x::b' i :a\ 
Now 2a represents the transverse axis AA'; and since x repre- 
sents AR, 2a + x will represent A'R ; therefore (2a+x)x rep- 
resents the product of the distances from the foot of the ordi- 
nate PR to the vertices of the curve. Hence we have the 
square of any ordinate to the transverse axis of an hyperbola, 
is to the product of its distances from the vertices of the curve, 
as the square of the conjugate axis is to the square of the 
transverse axis. 

If we draw a second ordinate P'R' to the transverse axis, we 
shall have 

PR 2 : AR x A'R : : V : a' : : P'R' 2 : AR' x A'R', 
or PR 2 : P'R' 2 : : AR x A'R : AR' x A'R' ; 

that is, the squares of ordinates to the transverse axis of an 
hyperbola are to each other as the products of the distances 
from the foot of each ordinate to the vertices of the curve. 



142 ANALYTICAL GEOMETRY. 

182. The equilateral hyperbola. When b is made equal to 
a, the equation of the hyperbola becomes 

y* = 2ax + x* (Art. 180), 
or tf=sf~-tf (Art. 170). 

The hyperbola represented by these equations is called equi- 
lateral or rectangular, and is to the common hyperbola what 
the circle is to the ellipse. 

Ex. 1. Trace the curve whose equation is y a = a? 2 — 16. 

Ex. 2. Trace the curve whose equation is y 2 =x 3 + 16. 

Ex. 3. Trace the curve whose equation is y 2 =zl0x+x\ 

183. To find the equation to the tangent at any point of an 
hyperbola. 

Let the equation to the hyperbola be a*y* — Va?=z—a*b\ 
Let x\ y r be the co-ordinates of the point on the curve at 
which the tangent is drawn, and #", y" the co-ordinates of an 
adjacent point on the curve. The equation to the secant line 
passing through the points x', y r and x'\ y" (Art. 40) is 

Eut, since the points x\ y r and x'\ y" are both on the hyperbola, 
we must have a 2 y /2 — & V 2 = — (tV, 

and a 2 y /,2 -b 2 x' /2 =-a 2 b 2 ; 

therefore, by subtraction, 

a*(y m r-y'*)-bXx m -x'*)=0 } 
y"-y' b\ x"+x' 

Substituting this value in equation (1), the equation of the se- 
cant line becomes 72 „ , * 

O X -f-X 

The secant will become a tangent when the two points coin- 
cide, in which case, x' — x n , and y f —y' r . 
Equation (2) will then become 

y-y'-jffi x ~ x ')> ( 3 ) 

which is the equation to a tangent at the point x\ y f . 



THE HYPERBOLA. 



143 



Clearing tills equation of fractions, and transposing, we ob- 
tain a?yy r — Vxx r = cfy" 1 — & V 2 ; 
hence a?yy' — Vxx' — —cfb*, (4) 
which is the simplest form of the equation to the tangent line. 

Vx f 
In equation (3), -7-7 represents the trigonometrical tangent 
a y 

of the angle which the tangent line makes with the transverse 

axis of the hyperbola. 



184. Points where the tangent cuts the axes. To determine 
the point in which the tangent intersects the 
axis of X, we make y=0, which gives 



a 



that is, x~ 

which is equal to CT. Therefore 
CTxCR=CA 2 . 
If from CR or x f we subtract CT, we shall 
have the subtangent 




KT=a>'- 



a 



x —a 



To determine the point in which the tangent intersects the 
axis of Y, we make x = y which gives 

b 2 

which is equal to Ct. Therefore Ct x 0=CB a . 

Hence it follows that if a tangent and ordinate be drawn 
from the same point of an hyperbola meeting either axis pro- 
duced, half of that axis will be a mean proportional between 
the distances of the two intersections from the centre. 

Ex. 1. In an hyperbola whose transverse axis is 32 inches, 
the abscissa of a certain point is 26 inches, and the ordinate 18 
inches, the origin being at the centre. Determine where the 
tangent passing through this point meets the two axes produced. 

Ex. 2. Find the angle which the tangent line in the preced- 
ing example makes with the axis of X. 



14-i ANALYTICAL GEOMETRY. 

185. To find the equation to the normal at any point of an 

hyperbola. The equation to a straight line passing through 
the point P, whose co-ordinates are x\y' (Art. 3S ). is 

y—tf=m(x—x : (1) 

and since the normal is perpendicular to the tangent, we shall 
have (Art ±6) 

1 
m— r» 

— m 

But we have found for the tangent line (Art. 1 v 

, W 

tn =-i— : 
ay ' 

hence n»= — j\ 

Substituting this value in equation (1), we shall have for the 
equation of the normal line 

where x and y are the general co-ordinates of the normal line, 
and x\ y the co-ordinates of the point of intersection with the 
hyperbola. 

186. Point with the axes. To find the 
point in which the normal cuts the tn 

Be axis, make y=0 in equation (2), and 
we have, after reduction. 

CJs =:»= j — . 

a 

If from CX we subtract CE. which is repre- 
sented bv x '. we shall have the subnormal 

To find the point in which the normal cuts the ax: 
make »=0 in equation rl\. and -we have, after reduction, 

which equals C 




THE HYPERBOLA. 



145 



187. Distance from {he focus to the foot of the normal. If 



we put c 1 for 



a* + l> 2 



a 



(Art. 177), we shall have 



CN=aV. 

If to this we add F'C (see next figure), which equals c or ac, 
Art. 177, we have 

F'N =zae+ eV = *(a + <?#') , 
which is the distance from the focus to the foot of the normal. 
Ex. In an hyperbola whose transverse axis is 32 inches, the 
abscissa of a certain point is 26 inches, and the ordinate IS 
inches, the origin being at the centre. Determine where the 
normal line passing through this point meets the two axes. 



188. A tangent to the hyperbola bisects the angle contained 
by lines drawn from the point of contact to the foci. 

Let PT be a tangent line to the hyperbola, and PF, PF' two 
lines drawn from the point of 
contact to the foci ; then the an- 



gle FPT=FTT. 

a' 



For CT = -(Art.lS4), 

X 

and CF=o*(Art.l77); 

hence ~FT = ae——=-(cx—a), 

a"* ft 

and F'T = ae+—=-(ex+a). 
x ar J 

Therefore 




FT:F'T::ea—a:ex+a, 

::PF:PF / (Art.l78). 
Hence PT bisects the angle FPF' (Geom.,Bk. IY., Prop. 17). 



189. If FT be produced to M, and the normal PN be drawn, 
it will bisect the exterior angle FPM. For, since PN is per- 
pendicular to TT', and the angle FPT is equal to FTT or its 
vertical angle MPT', therefore the angle FPN=MPjST; or the 
normal bisects the angle included by one radius vector and 

the other produced. 

G 



146 



ANALYTICAL GEOMETRY. 




190. To draw a tangent to the 
hyperbola through a given point 
of the curve. Let P be the given 
point ; draw the radii vectores PF, 
PF' ; on PF' take PG equal to PF, 
and draw FGL Draw PT perpen- 
dicular to FG, and it will be the 
tangent required, for it bisects the 
angle FPF ; . 

191. Every diameter of an hyperbola is bisected at the cen- 
tre. Let PP' be a straight line drawn through the centre of 
the hyperbola, and terminated on both sides by the two branch- 
es of the curve; it will be divided into two equal parts at the 
point C. Let x\ y f be the co-ordinates of the point P, and 
x", y n those of the point P'. 

Since the points P and P' are on the curve, we shall have 
(Art. 170) v 




and 



y"=-,(x"-a>), 
y'"Ax'"-ay, 



B. ~ a 

whence, by division, 



y 



x"-a' 



r 



lar, 



But, since the right-angled triangles CPU, CP'K' are similar, 
we have »/ x ' 






X 



hence, 



x 



-x' n -a? 



Clearing of fractions, we obtain 

x"=x" 2 ; 
whence also we have y f,1 =iy ,n . 

Consequently, x" + y" =x'"+ y n \ 

or CP 2 = CP' 2 ; 

that is, CP=CP'; 

that is, PP' is bisected in C. 



THE HYPEKBOLA. 



147 




192. Tangents to an hyperbola at the extremities of a diam- 
eter are parallel to each other. 

Let PP' be a diameter of an hy- 
perbola, and let PT, P'T' be tangents 
drawn through its extremities ; then 
is PT parallel to P'T'. 

In Art, 184 we found CT = ^,and 

a? 

for the same reason CT' = — , w T here x T represents CR, the ab- 

x 

scissa of the point P, and x" represents CE/, the abscissa of the 
point P'. But in Art. 191 we have found that x' —x" \ hence 
JT = CT'. The two triangles CPT, CP'T' have therefore two 
sides and the included angle of the one equal to tw r o sides and 
the included angle of the other; hence the angle CPT = the 
angle CP'T', and PT is parallel to P'T'. 



193. If from the extremities of the transverse axis two lines 
be drawn to meet on the curve, the product of the tangents of 
the angles which they form with that axis on the same side is 

equal to —. 

Let PA, PA' be two lines draw r n from the extremities of 
the transverse axis to the same point 
P on the hyperbola. The equation 
of the line PA passing through the 
point A, w T hose co-ordinates are x' ' = #, 
y' = 0,Ai± 38, is 

y—m{x— a). 
The equation of PA' passing through 
the point A', whose co-ordinates are x' r =. — a, 2/" = 0, is 

y—m\x-\-a). 
At the point of intersection, P, these equations are simultane- 
ous, and, combining them together, we have 

y'-^mm'^—a?). (1) 




148 ANALYTICAL GEOMETRY. 

But, since the point P is on the curve, we must have at the 
same time 

^=^-a 2 )(Art.l70). (2) 

Comparing equations (1) and (2), we see that 

mm ——, , 

where m denotes the tangent of the angle PAX, and ml de- 
notes the tangent of the angle PA'X. 

194. Definition. Two lines drawn from any point on the 
curve to the extremities of a diameter, are called supplement- 
ary chords. 

195. Supplementary chords in the equilateral hyperbola. 
In the equilateral hyperbola a—b, and we have 

?n??i' = l, 

1 



or m = —, 



f) 



m 

which shows that the angles formed by the supplementary 
chords with the transverse axis on the same side are comple- 
mentary to each other (Trig., Art. 28). 

196. If through one extremity of the transverse axis, a chord 
be drawn parallel to a tangent line to the curve, the supple- 
mentary chord will be parallel to the diameter drawn through 
the point of contact, and conversely. 

Let DT be a tangent to the hy- 
perbola at the point D, and let the 
chord AP be drawn parallel to it ; 
then will the supplementary chord 
AT be parallel to the diameter 
DC, which passes through the 
point of contact, D. 

Let x f , y' denote the co-ordinates 
of D. The equation of the line 
CD (Art. 31) gives y' = m'x ; 




THE HYPERBOLA. 149 

whence m —— . 

x 

But the tangent of the anMe which the tangent line makes 

■with the transverse axis (Art. 183) is 

m—-^—,. 
'-'■!/ 
Multiplying together the values of m and ?n', we obtain 



mm =-= 



i] 



cc 

which represents the product of the tangents of the angles 
which the lines CD and DT make with the transverse axis. 
But by Art. 193 the product of the tangents of the angles 

7 2 

PAX, PA X is also equal to -*. Hence, if AP be parallel to 
DT, AT will be parallel to CD, and conversely. 

197. The last Proposition is also true when applied to a tan- 
gent to the conjugate hyperbola; that is, if through one ex- 
tremity of the transverse axis' of an hyperbola a chord be 
drawn parallel to a tangent line to the conjugate hyperbola, 
the supplementary chord will be parallel to the- diameter 
drawn through the point of contact, and conversely. 

Let ET' be a tangent to the hyperbola which is conjugate to 
the former hyperbola, and let the 
chord AT be drawn parallel to ET', 
and through the point of contact, E, 
let the diameter EE' be drawn ; then 
will EE' be parallel to the supple- 
mentary chord AP. 

Let x", y" denote the co-ordinates 
of E. The equation of the line CE, 
Art. 31, gives y"=m'V; 

whence m"=?-r,. 

x 

The equation of the conjugate hyperbola (Art. 179) is 




150 ANALYTICAL GEOMETRY. 

and, proceeding as in Art. 183, we shall find that the tangent 
of the angle ET'C is 



7)1-- 



a y 



b* 
Hence we have mm"=—z* 

cr 

which represents the product of the tangents of the angles 

ECA and ETA.. But this has been found (Art. 196) equal to 

the product of the tangents of the angles DCX and DTX, or 

PAX. Hence, if AT be parallel to ET', AP will be parallel 

to CE, and conversely. 

198. Conjugate diameters. Each of the diameters DD', EE' 
is thus seen to be parallel to a tangent line drawn through the 
vertex of the other diameter. Two diameters thus related are 
said to be conjugate to each other. Thus we see that the pro- 
duct of the tangents of the angles which conjugate diameters 

. V 

form with the transverse axis is equal to —. 

199. Of any two conjugate diameters, one meets the original 
hyperbola, and the other the conjugate hyperbola. 

Let y=mx be the equation to any diameter, and let 
ay-bW^-a'b 2 
be the equation to the hyperbola. 

To determine the points in which the diameter intersects the 
curve, we must combine these two equations, and we have 

or # a = ji i— s- (1) 

b —am v J 

In like manner, for the conjugate hyperbola we shall find 

rf— ^- (2) 

x -tfm'-b 1 ' W 

The values of x in equation (1) will be real as long as a 3 m 2 
is less than Z> 2 , but imaginary when a^m 2 is greater than b\ In 
the former case the diameter intersects the curve, but in the 



THE HYPERBOLA. 



151 



latter it does not. The values of x in equation (2) are real 
when b 2 is less than a 2 m 2 , but imaginary when b 2 is greater 
than a 2 m 2 . 

Now, in the case of conjugate diameters, we have 

, v , ,. v 

mm — — , or m?n = — . 

J 2 £ 2 

Hence, if m 2 be less than — , m' 2 will be greater than — ; in this 

CO (t> 

case the first diameter meets the original hyperbola, and the 

b 2 

second the conjugate hyperbola. If m 2 be greater than — , m' 2 

b 2 a 

will be less than — : in this case the first diameter meets the 
a ' 

conjugate hyperbola, and the second the original hyperbola. 

200. Having given the co-ordinates of one extremity of a 
diameter •, to find those of either extremity of the diameter 
conjugate to it. 

Let AA', BB 7 be the axes of an hy- 
perbola, DD', EE 7 a pair of conjugate 
diameters. Let x\ y' be the co-ordi- 
nates of D ; then the equation to CD 
(Art. 40) is 



V 

y—— . x. 
u x 



(i) 




Since the conjugate diameter EE 7 is parallel to the tangent 
at D, the equation to EE' (Art. 183) is 



Li o / • X» 

ay 



(2) 



To determine the co-ordinates of E and E', we must combine 
the equation to EE 7 with the equation to the conjugate hyper- 
bola a 2 y 2 -b 2 x 2 =a 2 b 2 (Art. 179). 

Substituting the value of y from equation (2), we have, after 



reduction, 
whence 



(Fx'*-a*y'*)tf=ay 2 ; 



x l 



or 



ay ay 
= ~tfb 2 ^~V 

x-~ b . 



152 ANALYTICAL GEOMETRY. 

Also, from equation (2) we have 



y-- 



bx!_ 
: a' 



which are the co-ordinates of the points E and E'. The ab- 
scissa of E is positive, and that of E' negative ; hence the upper 
sign applies to E, and the lower to E'. 

201. The difference of the squares of any two conjugate di- 
ameters is equal to the difference of the squares of the axes. 

Let x\ y f be the co-ordinates of J) ; then, by Art. 200, 

CD'-CEW+y"-^— A- 
J o a 

FaT-cfy" a t y n -l i x n 

V + a? 

202. The rectangle contained by the focal distances of any 
point on the hyperbola is equal to the square of half the cor- 
responding conjugate diameter. 

LetDD', EE'be a pair of conju- 
gate diameters, and from D draw 
lines to the foci F and F'; then 
DFxDF' = CE 5 . 
Represent the co-ordinates of D 
by x', y'. 

Then, since CD 1 - 01?= a 1 - J 1 
(Art 201), we have 
CE»=CD'— a'+J* 




=x n +-{x n -a?)-a? + V 



-(i+j^-rf 



=«■- a 1 (Art 177) 
=DFxDF£Lrtl78); 

that is, the product of the focal distances DF, DF r is equal to 









THE HYPERBOLA. 



153 



the square of half EE', which is the diameter conjugate to that 
which passes through the point D. 

203. The parallelogram formed by drawing tangents 
through the vertices of two conjugate diameters is equal to 
the rectangle of the axes. 

Let DD', EE' be two conjugate diameters, and let DED'E' 
be a parallelogram formed by drawing 
tangents to the hyperbola through the ex- 
tremities of these diameters ; the area of 
the parallelogram is equal to AA' x BB 7 . 

Draw DM perpendicular to EE', and 
let the co-ordinates of D be x\ y'. 

The area of the parallelogram DED'E' 
is equal to 4CE . DM, which is equal to 4CE . CT sin. CTIT, 
which is equal to 4CT . EN", because EC and DT are parallel. 

(Art. 200). Hence the 




a* 



lx' 



a 



But CT=-7 (Art. 184), and EN: 

X a* bx' 

parallelogram DED'E' = 4^ . — =±ab = AA' x BB'. 

x a 



204. Equation to the hyperbola referred to any. two conju- 
gate diameters as axes. 

Let CD, CE be two con j ugate semi-diameters ; take CD as 
the new axis of x 9 CE as that of y ; let 
DC A = a, and EC A = j3. Let x, y be the 
co-ordinates of any point of the hyperbola 
referred to the original axes, and x', y r the 
cordinates of the same point referred to 
the new axes. 

The equation of the hyperbola referred to its centre and 
axes (Art. 170) is ay - 1 V = - a' b\ 

In order to pass from rectangular to oblique co-ordinates, 

the origin remaining the same, we must substitute for x and y 

in the equation of the curve (Art. 56) the values 

x—x f cos. a+y' cos. )3, 

y — x f sin. a + y' sin. )3. 

G2 




154 ANALYTICAL GEOMETRY. 

Squaring these values of x and y, and stflbstituting in the 
equation of the hyperbola, we have 

x'\c? sin. 2 a-6 2 cos.'a) + y'V sin. 2 /3-6 2 cos. 2 /3) 
+ 2x'y\a? sin. a sin. j3 — V cos. a cos. |3)= — & 2 6 2 , 
which is the equation of the hyperbola when the oblique co- 
ordinates make any angles a, j3 with the transverse axis. 
But, since CD, CE are conjugate semidiameters, we must 

have (Art. 198) , „ V 

mm = tang, a tang. j3 = — , 

whence a? tang, a tang. ]3 — J 2 = 0. 

Multiplying by cos. a cos. j3, remembering that cos. a tang, a 

=sin. a, we have 

a 2 sin. a sin. j3— V cos. a cos. j3 = 0. 
Hence the term containing x f y r vanishes, and the equation be- 
comes 

x'\a % sin. 2 a-& 2 cos. 2 a) + 2/V sin. 2 /3-5 2 cos. 2 j3) = -a 2 6 2 , 
which is the equation of the hyperbola referred to conjugate 
diameters. 

If in this equation we suppose y f — 0^ we shall have 

*, ^ 

b COS. a — a sin. a 
This is the value of CD 2 , which we shall denote by a' 2 . 
If we suppose x' = 0, we shall have 

Now, since we have supposed that the new axis of x meets the 
curve, we know that the new axis of y will not meet the curve 
(Art 199), so that —a'b' 

a'sm. a /3-^cos. s /3 
is not a positive quantity. If we denote it hy — J' 5 , the equa- 
tion to the hyperbola referred to conjugate diameters will be 

b n x n -ay=a ,t b n i 
or, suppressing the accents on the variables, 






THE HYPERBOLA. 



155 



205. The square of any diameter of an hyperbola is to the 
square of its conjugate, as the rectangle under any two seg- 
ments of the diameter is to the square of the corresponding 
ordinate. 

Let DD', EE' be two conjugate diameters of an hyperbola, 
and from any point of the curve, as 
P, let PR be drawn parallel to EC, 
meeting the diameter DD' produced 
iiiR. 

The equation of the hyperbola re- 
ferred to conjugate diameters may 
be put under the form 

a , Y=5 ,, (« 9 --a ,f ). 
This equation may be reduced to the proportion 




a! % \V*\\tf—a n 



w 



or (2a'Y : (2bJ : : (x+a'){x-a r ) : y\ 

Now 2a' and 2b' represent the conjugate diameters DD',EE'; 
and since x represents CE, x+a' will represent D'R, and x—a' 
will represent DP ; also PR represents y. Hence 
DD /2 :EE /2 ::DRxRD':PR\ 
If we draw a second ordinate P'R' to the diameter DD 7 , we 
shall have PR 2 : DR x RD' : : b" :a' 2 :: P'R /2 : DR' x RT)', 
or PR 2 : PR /2 : : DR x RD' : DR' x R'D' ; 

that is, the squares of any two ordinates to the same diameter 
are proportional to the rectangles under the corresponding 
segments of the diameter. 

206. To find the polar equation to the hyperbola, the pole 
being at one of the foci. 
1. Let F be the pole. 
Let FP=?7 the angle AFP=0; 
then FR=r cos. PFR= — r cos. 0. 
By Art. 178, 

r=ex— a ; 
but a>=CR=CF+FR 

=ae—r cos. 0. 




15 G ANALYTICAL GEOMETRY. 

Therefore r=ae*— er cos. — a. 

Hence r(l + e cos. 0) = a(e* — 1), 

or r^-^ m (!) 

1 + e cos. 0' x J 

which is the equation required. 

2. Let F' be the pole. 

Let FP=r7 angle PF'A=0'; then F'R=/ cos. 0'. 

By Art. 178, 7*' = <saj+0; 

but a=CR=F'R-F'C 

=7*' cos. Q'—ae. 
Therefore r' — er' cos. 0' — a<? 2 + a. 

Hence r'(l - e cos. 0') = a(l - e*) = -afc 1 - 1), 

or r'=^ L (2) 

1 — e cos. ' v y 

which is the equation required. 

207. Form of the hyperbola traced. The form of the hy- 
perbola may be traced from its polar equation. In equation 
(1), suppose 0=0; then r=a(e—l). If we measure off this 
length on the initial line from the pole F, we shall obtain the 
point A as one of the points of the curve. 

While increases, 1 + e cos. diminishes, and r increases ; 

V 
and when = 90°, r=—, which determines another point of 

the curve. 

When becomes greater than 90°, cos. becomes negative, 

and r continues to increase until 1 + e cos. = 0, or cos. 0= — -, 

e' 

when r becomes infinite. Thus, while increases from until 
cos. 0=—-, that portion of the curve is traced out which be- 
gins at A, and passes on through P to an indefinite distance 
from the origin. 

When 1 + e cos. becomes negative, r becomes negative, and 
we measure it in the direction opposite to that in which we 
should measure it, if it were positive. Thus, while increases 



THE HYPERBOLA. 



157 



to 1S0°, that portion of the curve is traced out which begins 
at an indefinite distance from C in the lower left-hand quad- 
rant, and passes through Q to A'. 

As increases from 180°, r continues negative, and increases 
numerically until 1 + e cos. again becomes zero. Thus the 
branch of the curve is traced out which begins at A', and pass- 
es on through Q' to an indefinitely great distance from C. 

As 6 continues to increase, the value oil + e cos. again be- 
comes positive ; r is again positive, and is at first indefinitely 
great, and then diminishes. Thus the portion of the curve is 
traced out which begins at an indefinitely great distance from 
C in the lower right-hand quadrant, and passes on through P' 
to A. Thus both branches of the hyperbola are traced out by 
one complete revolution of the radius vector. 



208. Any chord which passes through the focus of an hyper- 
bola is a third proportional to the transverse axis and the di- 
ameter parallel to that chord. 

Let PP' be a chord of an hyperbola passing through the 
focus F, and let EE' be a diameter par- 
allel to PP'. 

By Art. 206, PF = .f" 1 ^ . 
J ' ± + e cos. 

To find the value of FP', we must sub- . 

stitute for 0, 180° + 0, and we obtain 

a(e 2 ^l) 



Hence 



FP': 



pp , 




1 — e cos. 0' 
2a(6 2 -l) 



1WCOB. 8 0' 

Proceeding as in Art. 161, we find the value of CE 2 equal to 



Hence 



pp/ 



1-£ 2 COS.'0' 

2CE 2 4CE 2 



a ~" 2a ' 
that is, AA':EE'::EE':PP', 

or PP' is a third proportional to AA' and EE'. 



158 ANALYTICAL GEOMETKY. 

209. Definition. The parameter of any diameter is a third 
proportional to that diameter and its conjugate. 

The parameter of the transverse axis is called the principal 

parameter, or latus rectum, and its value is — (Art. 176). The 

parameter of the conjugate axis is —r-. The latus rectum is 

the double ordinate to the transverse axis passing through the 
focus (Art. 176). Now, since any focal chord is a third propor- 
tional to the transverse axis and the diameter parallel to that 
chord, and since the transverse axis is less than any other di- 
ameter of the same hyperbola, it follows that the transverse 
axis is the only diameter whose parameter is equal to the 
double ordinate passing through the focus. 

In the equilateral hyperbola a= b, and the latus rectum is 
equal to either of the axes of the curve. 

210. Definition. The directrix of an hyperbola is a straight 
line perpendicular to the transverse axis, and intersecting it in 
the same point with the tangent to the curve at one extremity 
of the latus rectum. 

Thus, if LT be a tangent drawn through one extremity of 
the latus rectum LI/, meeting the transverse axis in T, and NT 
be drawn through the point of intersection perpendicular to 
the axis, it will be the directrix of the hyperbola. 

The hyperbola has two directrices, one corresponding to the 
focus F, and the other to the focus F\ 

211. The distance of any point in an hyperbola from either 
focus is to its distance from the corre- 
sponding directrix, as the eccentricity 
is to unity. 

Let F be one focus of an hyperbola, 
NT the corresponding directrix ; F' the 
other focus, and N'T' the corresponding 
directrix. Let P be any point on the 
hyperbola, x' ', y r its co-ordinates, the origin being at the centre. 




THE HYPERBOLA. 159 

Join PF, PF', and draw PNN' parallel to the transverse axis, 
and PR perpendicular to it. 

By Art. 184, CT=-=-; 

C €> 

hence CK-CT = PN=£'-- 

e 

ex'— a 



e 
But, by Art.173, r=ex'-a=FF; 
hence <?.PN=PF, 

or PF:PN::*:1. 

In like manner we find that 

PF':PN'::*:1. 

212. Conic sections compared. In Art. 82 the parabola was 
defined to be a curve every point of which is equally distant 
from the focus and directrix, while in the ellipse and hyper- 
bola these distances have been found to be in the ratio of the 
eccentricity to unity. In the ellipse, the eccentricity, being 

equal to - (Art. 127), is less than unity, while in the hyperbola 

Cb 

(Art. 177) it is greater than unity. In each of these curves the 
two distances have to each other a constant ratio. In the par- 
abola this ratio is unity, in the ellipse it is less than unity, while 
in the hyperbola it is greater than unity. These curves, being 
the sections of a cone made by a plane in different positions, 
are called the conic sections ; so that a conic section may be 
defined to be a curve traced out by a point which moves in 
such a manner that its distance from a fixed point bears a con- 
stant ratio to its distance from a fixed straight line. If this 
ratio be unity ^ the curve is called a parabola ; if less than unity, 
an ellipse ; and if greater than unity, an hyperbola ; and all 
the properties of these curves may be deduced from this defi- 
nition. 



ICO 



ANALYTICAL GEOMETRY. 



ON THE ASYMPTOTES OF THE HYPERBOLA. 

213. It was shown in Art. 199 that if a line drawn through 
the centre of an hyperbola meets the curve, m a must be less 

7 3 Z 

than — , or m<±-/ and if the line meets the conjugate hy- 
perbola, m 2 must be greater than — , or ra> ±-. 

Let AA'j BB' be the two axes of an hyperbola, and through 

p^ L the vertices A, A', B, B' let lines be drawn 

perpendicular to these axes; and letDD', 

EE', the diagonals of the rectangle thus 

formed, be indefinitely produced. 

Then, since -^ A 7 

^^^. DA b 
tans;. DCX=-7-tt=-, 

to AC a? 

tang.E , CX=^=-A 

° AC a? 

it follows that the lines CD, CE' will never meet the curve at 
any finite distance from C. 

The lines CD, CE', indefinitely produced, are called asymp- 
totes of the hyperbola. 




and 



214. Definition. An asymptote of any curve is a line which 
continually approaches the curve, coming indefinitely near to 
it, but meets it only at an infinite distance from the origin. 

Since the lines DD' and EE' pass through the centre, and 
are inclined to the transverse axis at an angle whose tangent 



:-, their equation will be 



h 

z-X. 

a 



215. The diagonals of the rectangle formed by lines drawn 
through the extremities of the axes and perpendicular to the 
axes, are asymptotes to the curve, according to the definition 
of Art. 214. 



THE IIYrERBOLA. 1G1 

Let the equation to the hyperbola (Art. 170) be 

The equation to the line CL, the diagonal of the rectangle 
DED'E',is bx 

J a 
Let MPR be an ordinate meeting the hyperbola in P, and the 
straight line CL in M ; then, if CR be denoted by x, we have 

PR=-V^W, 

a 

bx 

and MK=— . 

7 a 

Hence MP =-(<*?- Vx u -d 2 ) 

b a 2 



a x + Vvf—cf 
ab 



x-{- Vx'—a? 

If, then, the line MR be supposed to move from A parallel 
to itself, the value of x will continually increase, and the dis- 
tance MP will continually diminish ; and if we suppose the 
point P of the curve to recede to an infinite distance from the 
origin, MP will become zero. * 

In like manner the line CL 7 , whose equation is y—~- — > 

meets the curve below the transverse axis at an infinite dis- 
tance from the origin. 

216. Asymptote to the conjugate hyperbola. The line CL is 
also an asymptote to the conjugate hyperbola; for, let PR be 
produced to meet the conjugate hyperbola in P'; then (Art. 179) 

P'R=-V^+^". 
a 

b. 



Hence P'U=-(Vtf + a*--x) 

a K 2 } 

aa 

Vx^+d' + x 



162 ANALYTICAL GEOMETEY. 

Theref ore, if CR or x be indefinitely increased, P'M will be 
indefinitely diminished, and hence CL is an asymptote to the 
conjugate hyperbola. 

217. An asymptote may be considered as a tangent to the 
hyperbola at a point infinitely distant from the centre. 

The equation to a tangent at any point a/, y f of the curve 

(Art. 183) is a?yy f - Vxxf = - a?b% 

Vxx' b 1 
or y—- 1 — r ——. (1) 

«y y 



Now y'z^dz-i/x'* 



a 



■a . 



If x r becomes indefinitely great, then a? vanishes when com- 
pared with a? /2 , and we have 7 



* a 
Substituting this value in equation (1), the equation to the 
tangent, when the point #', y' is infinitely distant, becomes 

Vxx' a ab 2 
y~~~ a 2 bx f ~~bx f 
bx ab 
a x 

But when x' is infinite, -7=0 ; 

h\rp 

hence V—^ — > 

J a J 

which is the equation to the asymptote (Art. 214). Hence the 

asymptote is a tangent to the curve at a point infinitely distant 

from the centre. 

218. The asymptotes are the diagonals of every parallelo- 
gram formed by drawing tangents through the vertices of two 
conjugate diameters. 

Let DED'E' be a parallelogram formed by drawing tangents 
to the hyperbola through the vertices of two conjugate diam- 
eters DE>', EE'; the diagonals Tt, TV will be asymptotes of 
the curve. 



THE HYPERBOLA. 163 

Let x\ y' be the co-ordinates of the point D ; then the co- 
ordinates of E, the extremity of the 
conjugate diameter (Art. 200), are 

ay' _ bx' 

-T- and — . 
o a 

Draw the diagonal DE, and it will 

bisect CT in N (Geom., Bk. L, Prop. 33). 

The co-ordinates of N are 




i(, + f) and i^). 



Hence we have 



y'+— h 

tang. ]STCX= „ which equals -. 

But - is the tangent of the angle which the asymptote makes 

with the transverse axis (Art. 214) ; hence CT coincides with 

one of the asymptotes. 

Also, since the diagonal DE passes through the points 

, , n ay' bx' 
#,2/',and-|-, — , 

the tangent of the angle which it makes with the transverse 
axis (Art. 40) is % x ' 



b h 



1. which equals — -. 
ay'' L a 



But —-is the tangent of the angle which the other asymptote 



a 



makes with the transverse axis ; hence DE is parallel to the 
other asymptote. And since DT 7 E 7 C is a parallelogram, DT 7 
=E'C, which equals EC ; and since DT 7 is parallel to EC, ED 
is parallel to CT 7 . Hence TV is the other asymptote. 

219. Hence we see that the line joining the extremities of 
two conjugate diameters is parallel to one asymptote, and is 
bisected by the other. 



164 ANALYTICAL GEOMETRY. 

Also, if a tangent line he drawn at any point of an hyper- 
bola, the part included between the asymptotes is equal to the 
parallel diameter. 

Moreover, if x and y are the co-ordinates of any point on the 
asymptote referred to two conjugate diameters, then we shall 
have * y : x : : V : a', 

Vx 

which is therefore the equation to the asymptote referred to a 
pair of conjugate diameters. 

220. If any chord of the hyperbola be produced to meet the 
asymptotes, the parts included between the curve and the as- 
ymptotes will be equal. 

Let PP' be any chord of the hyperbola, and let it be pro- 
duced to meet the asymptotes in M and M' ; 
then will PM be equal to P'M'. 

Draw CY, the semidiameter to the conju- 
gate hyperbola, parallel to PP', and draw CX 
conjugate to CY ; then PP' is a double ordi- 
nate to CX, and is bisected in R. 

The equation to the hyperbola referred to 
CX, CY (Art. 204) is 




y=±jVtf=*% (1) 

and the equation to the asymptotes (Art. 219) is 

b r 

y=±^ x - ( 2 ) 

Now to the same abscissa CR there correspond (from eq. 1) 
two equal ordinates w T ith opposite signs ; hence we have 

PR=P'R. 
Also, from eq. 2, MR=M'R. 

Therefore, by subtraction, MP = MT', 
as was to be proved. 

If the tangent line TT' be drawn parallel to MM', the trian- 
gles CTT, CMM' will be similar ; and since MR is equal to 



THE HYPERBOLA. 



1G5 



M'R, NT will be equal to NT' ; that is, the portion of a tan- 
gent included between the asymptotes is bisected at the point 
of contact. 

221. If a straight line be drawn through any point on an 
hyperbola, the rectangle of the parts intercepted between that 
point and the asymptotes, will be equal to the square of the 
parallel semidiameter. 

Let a straight line drawn through the point P on the hyper- 
bola meet the asymptotes in M and M'; then we have 
PM . PM / = (MR~PE)(MR+f > K) 
=MR 2 -PR 2 . 



But 



and 



hence 



MK'=— ]a? (Art. 219), 
YR % ~(a?-a'*) (Art. 204) ; 



MR*- PR': 

a 

that is, PM.PM , = 5 / ", 

or, the rectangle of the parts PM and PM' is equal to the 

square of the parallel semidiameter. 

222. To find the equation to the hyperbola referred to the 
asymptotes as axes. 

Let CX, CY be the original axes coinciding with the axes of 
the hyperbola, and let CD, CE be the 
new axes, inclined to CX on opposite 
sides of it at an angle ]3, such that 

tang. ]3 = - (Art. 213). Let x, y be the 
a 

co-ordinates of a point P referred to the 
old axes, and x' ', y r the co-ordinates of 
the same point referred to the new axes. 

The formulas for passing from rectangular to oblique co-or- 
dinates, the origin remaining the same (Art. 56), are 

x—x' cos. a+y f cos. /3, 

y—x' sin. a + y' sin. /3. 




166 ANALYTICAL GEOMETRY. 

But, since a= — /3, these equations become 
x=(x'+y') cos./3, 
2/=(^ / — 2/0 sin. j3. 

Now sin. /3 = -7tf> and cos. p = j^r ; 

also, CL a = C A a + AL a = a 9 + b\ 

Represent CLhjc; then 

sin. /3 = -, and cos. 8 = -. 

_ . | a(x'+y') b(x'-y') 
Therefore x = ~ — . and y = — . 

Substitute these values in the equation to the hyperbola, 

ay-Vx^-aW, 
and we have 

dl\x> - yj - a*b\x' + yj = - aW, 

or W-tff-W+ifT=-*i 

that is, 4aj'y'=c"; 

or, suppressing the accents, 

c a a a + & a 

which is the equation of the hyperbola referred to the asymp- 
totes as axes. 

223. Equation to the conjugate hyperbola. The equation 
to the conjugate hyperbola referred to the same axes may be 
found by writing — a? for & a , and — V for b 2 (Art. 179). We 
shall th§a have 

a' + b* 

In the case of an equilateral hy perbola, the angle DCE = 90° ; 
that is, the asymptotes are perpendicular to each other. For 
all other hyperbolas the asymptotes make oblique angles with 
each other. 

Ex.1. Trace the curve whose equation referred to rectangu- 
lar axes is xy=10. 






THE HYPERBOLA. 



167 



We may assume any value for x, and the corresponding value 
of y may be found from the equation. Thus, if 



<y-- 



x-- 

x=2,y= 

es=S,y= 

x=4,y= 

x=5,y= 

x=6,y= 



:10. 



5. 

3.33. 

2.5. 

i. 

1.66. 



x= 7,y=lA3. 
x= 8,y=1.25. 
x- 9,y=l.ll. 
aj=10,y=1.00. 
SB=ll,y=0.91. 
aj=12,y=0.83. 



\ 



Tfrrrr . 



These values determine the points of the curves a, b, c, d, etc. 
If x is negative, y is also negative, 
and the points a' ', b\ </, etc., will be 
determined in the third quadrant. 
As x increases indefinitely, y de- 
creases, and the curve is unlimited m I Jl ' I r m 
in the direction of x positive, but *dH 

continually approaches the axis of 
x without actually reaching it. The 
same is true for the direction of x 
negative, and for each direction of the axis of y. 

Ex. 2. Trace the curve whose equation referred to oblique 
axes is xy= — 10. 



224. Parallelogram on any abscissa and ordinate. 
be any point on the hyperbola, from which draw 
PM, PjS" parallel to the asymptotes, and repre- 
sent these co-ordinates by a?, y ; then, by Art. 



LetP 




K 



If we multiply each member of this equation by sin. 2/3, we 
shall have 

xy sin. 2/3 = °—^- sin. 2/3, (1) 

where 2/3 is the angle included by the asymptotes. The first 
member of this equation represents 

CMxCNx^n.MCK (2) 

But CX x sin. MCN is the perpendicular from N upon the line 



168 ANALYTICAL GEOMETRY. 

CM ; hence expression (2) represents the area of the parallelo- 
gram CNPM. 

Since sin. 2/3=2 sin. ]3 cos. /3=^y 2 (Art. 222), 

the second member of eq.l reduces to -^. 

Hence the parallelogram CNPM described on the abscissa 
and ordinate of any point on the curve, is equal to half the 
rectangle under the semiaxes, or one eighth the rectangle under 
the axes. 

225. To find the equation to the tangent at any point of an 
hyperbola when the curve is referred to its asymptotes as axes. 

Let x\ y r be the co-ordinates of the point of contact, x", y" 
the co-ordinates of an adjacent point on the curve. 

The equation to the secant line passing through these points is 

y-y'=wEi^- x ')- CO 

Since the two given points are on the hyperbola, we have 

(Art. 222) 

(? 
xy 



"4 



x u 
Hence x'y' '—x ff y'' ', or y" 



fr n ,tt 



X 



x'lf 

Therefore y"— y' =-77— y 



x 



whence 



y -y y_ 

x"-x'~~x" 
By substitution, eq. (1) becomes 



y-y'^-^rix-x'). (2) 






If we suppose #'=#", smd-y'—y", the secant will become a 
tangent, and equation (2) will become 






THE JIYrEEBOLA. 



100 



y-y 



: ,(x-x% 



which is the equation to the tangent line. 

If we clear this equation of fractions, we shall have 
yx'-x'y'=-xy' + x'y'; 

ff 1 I 7 2 

therefore yx' + xy' — 2x'y' = — - — ,* 

which is the simplest form of the equation to the tangent line. 

226. Points of intersection with the axes. To find where 
the tangent at x\ y f meets the axis of ab- 
scissas, put y — in the equation to the tan- 
gent line, and we have 

or x=2x / ; 

that is, the abscissa CT' of the point where 
the tangent meets the asymptote CE is 
double the abscissa CR of the point of tangency. 

To find where the tangent cuts the axis of Y, put x=0 in the 
equation to the tangent line, and we have 

ijx' = 2xy\ 
or y = 2?j; 

that is, CT is double of PR. 

Also, because PR is parallel to CT, TT' is double of PT, or 
the tangent TT' is bisected in P; that is, if a tangent line he 
drawn at any point of an hyperbola, the part intercepted be- 
tween the asymptotes is bisected at the point of contact. 

II 



eTt 7 " 



170 ANALYTICAL GEOMETRY. 



SECTION VIII. 

GENERAL EQUATION OF THE SECOND DEGREE. 

227. We have seen that the equations of the circle, the par- 
abola, ellipse, and hyperbola are all of the second degree ; we 
will now inquire whether any other curve is included in the 
general equation of the second degree. 

The general equation of the second degree between two va- 
riables may be written 

ax* + bxy + cif + dx + ey -\-f= 0, (1) 

which contains the first and second powers of each variable, 
their product, and an absolute term. 

We shall suppose the axes to be rectangular ; for if they were 
oblique we might transform the equation to one referred to 
rectangular axes, and we should obtain an equation of the same 
degree as the above, and which could not, therefore, be more 
general than the one we have assumed. 

228. To remove certain terms from the general equation. 
We wish, if possible, to cause certain terms of this equation to 
disappear. For this purpose we may change both the origin 
and direction of the co-ordinate axes, without assigning any 
particular values to the quantities which determine the position 
of the new axes. By this means, indeterminate quantities are 
introduced into the transformed equation, to which such values 
can afterwards be assigned as will cause certain of its terms to 
vanish. Instead of chan^inc; both the origin and direction of 
the co-ordinate axes at once, it is more convenient to effect 
these changes successively. 

229. The terms containing the first powers ofx and y in 
the equation of the second degree, may in general he made to 
disappear by changing the origin of the co-ordinates. 



GENERAL EQUATION OF THE SECOND DEGREE. 171 

In order to effect this transformation, substitute for x and y 
in equation (1) the values 

x=zx'+h, 

by which we pass from one system of axes to another system 
parallel to the first (Art. 54). 
The result of tliis substitution is 

ax' 2 + bx'y' + erf 1 + (2ah + bk + d)x' + (2ck + bh + e)y' 
+ at? + bhk + d? + dh + ek +f= 0. 
Now, in order that the terms involving the first powers of x' 
and y' may disappear, we must have 

2ah + bk±d=0, \ 

and 2c&+bh+e = Q. 

From these equations we obtain 

2cd—be 2ae—bd 

*=■?=& and ^="^=i^- 

These are the values of h and k which render the proposed 
transformation possible ; hence, denoting the constant quantity 

ah* + bhk+ck 2 + dh + ek+f 
byf, the transformed equation becomes 

ax" + bx'y' + cy" +/' = 0. (2) 

When b 2 —4:ac—0 y the above values of h and k become infi- 
nitely great, and the proposed transformation is impossible. 

If equation (2) is satisfied by any values x v y l of the varia- 
bles, it is also satisfied by the values — & v —y v Hence the new 
origin of co-ordinates is the centre of the curve represented by 
equation (1). 

Thus, if U — kac be not =0, the curve represented by (1) has 
a centre^ and its co-ordinates are h and k, the values of which 
are given above. 

We may suppress the accents on the variables in equation 
(2), and write it ax 2 + bxy + cif +/' = 0. (3) 

230. The term containing xy in the general equation of the 
second degree may be taken away by changing the directions 
of the axes. 



172 ANALYTICAL GEOMETRY. 

For this purpose put 

x—x f cos. 0—y' sin. 0, 
y—x f sin. 9+y f cos. 6. 
Substituting these values of x and y in equation (3), and ar- 
ranging the result, we have 

x f \a cos. 2 + c sin. 2 + £ sin. 6 cos. 0) 
+y f \a sin. 2 + <? cos. 2 0— b sin. cds. 0) 
-f#y{2(<?— a) sin.0 cos. + %os. 2 0-sin. 2 0)}+/ / = O. (4) 
Now, in order that the term involving x'y f may become zero, 
we must have 

2(c— a) sin. cos. + 5(cos. 2 0-sin. 2 0) = O. 
But by Trig., Art. 73, 

2 sin. cos. = sin. 20; also cos. 2 0— sin. 2 = cos. 20; 
hence (c— a) sin. 20 + 5 cos. 20 = 0, 

or tang. 20=^. (5) 

Since the tangent of an angle may have any magnitude from 
zero to infinity, this value of tang. 20 is always possible, what- 
ever be the values of a, £, and c ; hence such a value of may 
always be found as shall remove the term involving x'y' from 
equation (4), and the general equation is reduced to the form 

AiB' a +By /a +/'==0, 
or, suppressing the accents on the variables, we have 

A0 9 +B^+/'=O. (G) 

By solving this equation we have 



. 



V 



-f'-W 



from which we see that if A, B, and/* 7 have the same sign, the 
quantity under the radical is negative, and equation (G) repre- 
sents an imaginary curve. 

If A and B have the same sign, and f have the contrary 
sign, the equation represents an ellipse (Art. 121). 

If A and B have different signs, the equation represents an 
hyperbola (Art. 170). 

If A =B, the equation represents a circle (Art. GO). 

If y = 0, and A and B have the same sign, the equation can 



GENERAL EQUATION OF THE SECOND DEGREE. 173 

only be satisfied by the values # = and y = 0; that is, the 
equation represents a point, viz., the origin. 

If f = 0, and A and B have different signs, equation (6) re- 

duces to y=i±.x\J — p, 

which represents two intersecting straight lines. 

231. To find the values of the coefficients A and B in equa- 
tion (6) in terms of 'a, b, and c. 

Since A = a cos. 2 + c sin. 2 + i sin. cos. 0, 
and B=a sin. 2 + <? cos. 2 0— £ sin. cos. 0, 

we have, by addition, observing that sin. 2 + cos. 2 = 1, 

A+B = a + <?, (m) 

and by subtraction, observing that cos. 2 0— sin. 2 = cos. 20, 
A-B=(a— c) cos. 20+ J sin. 20. 
JSTow, since sec. = Vl + tang. 2 , 



byeq.(5), sec. 20 = ^4 *' V ^~ C) ' +b 



hence cos. 20: 



{a—cf a—c 

cc—c 



and sin. 20— , =. 

Vu+(CL-Cf 

Hence we have 

(a-cy 



A-B = 



V+(a-cy / 

= V y+(«- c y ==± ^ , +(«- c )'- <*> 

Adding and subtracting successively (m) and (»), we have 

B=l{a-c+W + (a-cy}. 
Multiplying together these values of A and B, we have 

. _, (a + cy-U-ia-cy Aac-V 

A . B = ; — -. . 

4 4 

Hence A and B have the same sign or different signs accord- 
ing as ^ac—V is positive or negative. 



174 ANALYTICAL GEOMETRY. 

232. Particular case considered. We will now consider 
the case in which b 2 — kac is zero. We can not in this case 
destroy the terms involving x and y by transferring the origin 
to the centre of the curve, as was done with the ellipse and 
hyperbola, but we may remove the term involving xy by chang- 
ing the direction of the axes. 
Let the equation be 

ax* + bxy + cy* +dx + ey +f— 0. (1) 

Put x — x f cos. 9— y' sin. 0, 

y=x ; sin. 9-\-y f cos. 0. 
Substituting these values in equation (1), we have 
x'\a cos. 2 + c sin. 2 + £ sin. cos. 0) 
+y r \a sin. 2 + <? cos. 2 — b sin. cos. 9) 
+ xy{2(c-a) sin. 9 cos. 0+5(eos. 2 0-sin. 2 0)} 
+x'(d cos. 9+e sin. 9)+y\e cos. 9—d sin. 0)+/=O. (2) 
In order that the term involving x f y' may become zero, we 
must have 2(c— a) sin. 9 cos. + £(cos. 2 0— sin. 2 0) = O; 

whence, as in Art. 230, tans;. 29 — , 

7 & a— <y 

and the co-efficients of x n and y'% as in Art. 231, are 

±{a + c+Vb* + (a-cy}. 

One of these coefficients must therefore vanish, since their prod- 

4:dC b* 

net (Art. 231) is —j — , which, by hypothesis, =0. Suppose 

the coefficient of x ;2 — ; if we suppress the accents on the va- 
riables, equation (2) will assume the form 

(Y + Da?+Ey+/=0. (3) 

Transposing and dividing by C, we have 

f E?/ T>x f 

E , y + c - c a 

Adding -r^ tb each member, we have 

/ EV D/ E 2 f\ 

t^ . 7 E ^ r D a E 2 / . . /( 

Tut £= — —-„ M=— p, and w = jfttt— ^k, and equation (4) 

may be written (y— Z) 2 — M(# — n). 






GENERAL EQUATION OF THE SECOND DEGREE. 175 

If now the origin be transferred to a point whose co-ordi- 
nates are x~n, y—l, 
we shall have, by writing x + ?i for x, and y + l for y, 

3f=M«, (5) 

which is the equation to a parabola. 
If in equation (3) D = 0,we have. 

Cy+Ey+/=0, 
"F 1 
which gives y= ""20^2(3 ^-lOf, 

which represents two parallel straight lines, or one straight line, 
or an imaginary curve, according as E 2 is greater, equal to, or 
less than 4tCf. 

233. Conclusions. Hence we arrive at the following results : 
The equation ax? + bxy + cy x + dx + ey +f=- 
represents an ellipse, if 6 2 — 4ac be negative, subject to three ex- 
ceptions, in which it represents respectively a circle, a point, 
and an imaginary curve (Art. 230). 

If V—kac be positive, the equation represents an hyperbola, 
subject to one exception when it represents two intersecting 
straight lines (Art. 230). 

If V— 4:ac=0, the equation represents a parabola, subject to 
three exceptions, in which it represents respectively two paral- 
lel straight lines, one straight line, and an imaginary curve 
(Art. 232). 

Ex. 1. Determine the form and situation of the curve repre- 
sented by the equation 

x 2 —xy+y* — 2x—2y+2 — 0. 

Here If— \ac—— 3; hence the equation represents an el- 
lipse (Art. 233). 

In order to transfer the origin to the centre of the curve, we 
substitute h+x' for x, and k + y' for y. The values of h and k 
are given by the formulas of Art. 229, 

7 -4-2 rt 1 -4-2 

also, /' = 4-4+4-4-4 + 2=-2. 



17G 



ANALYTICAL GEOMETRY. 



Hence the transformed equation is 

Next, retaining the centre of the ellipse as the origin, we 
must find through what angle the axes must be turned in order 
that the term containing xy may vanish. 

By Art. 230, tang. 20 = —— = = jj~ = infinity; hence 20 = 90°, 

and 0=45°. 

Also, by Art, 231, A=4(2 + VT)=f, 

and B=-i(2-Vl)=4. 

Therefore the equation to the ellipse referred to the new axes is 



x 4. 



% 2 



2 -+-f--2=0, 

or ^ 2 + 3t/ 2 = 4. 

2 4 

The semiaxes are —n\ and 2, and the axes are —^ and 4. 

The annexed figure represents the form of the curve, and 
its position with respect to the different 
/X systems of axes, the co-ordinates of A' 
being (2, 2), and the angle X'A'X" be- 
ing 45°. 

x 1 — xy+?/-2x-2y+2 = 
is the equation of the ellipse referred 
to the axes AX, AY. 

x 12 — xy+if — 2 = 
is the equation of the same ellipse referred to the axes A'X', 
AT. 

# 2 + 3// 2 = 4 
is the equation of the same ellipse referred to the axes A'X", 
A'Y". 

Ex. 2. Determine the form and situation of the curve repre- 
sented by the equation 

x 2 — Gxy + y'-Gx + 2y + 5 = 0. 
Here V — 4a<? = 3G — 4^32 ; hence the equation represents 
an hyperbola. 

By the formulas of Art. 229 we find 



. 




GEXEEAL EQUATION OF THE SECOND DEGREE. 177 



/;=l-2+5=4. 

Hence, when the origin is transferred to the point (0, —1), the 
equation becomes x*— 6xy+y* + 4z = 0. 

In order that the term containing xy may vanish, we must 

have tang. 20 = — { y = infinity. Hence 0=45°. 

Also, A = ^(2+ V36)=+4, 

and B=|(2-.V36)=-2. 

Hence the transformed equation is 

4?/ 2 -2^ + 4 = 0. 
The student should construct a figure showing the form and 
position of the curve with respect to the different axes of ref- 
erence. 

Ex. 3. Determine the form and situation of the curve repre- 
sented by the equation 

x*-2xy + y 2 -8x+lQ = 0. (1) 

Here i 2 — Aac=0 ; hence the equation represents a parabola. 
Substituting for x in eq. (1), 

x' cos. 9—y f sin. 0, 
and for y, x' sin. 9-\-y r cos. 0, 

we obtain an equation of the form 

Ax'+Bxy+Ctf + Dx + Ey + F^O, (2) 

where A = 1 — sin. 20, D = — 8 cos. 9, 

B = -2(cos. 2 0-sin. 2 0), E= 8 sin. ft 
C= 1 + sin. 20, F=16- 

Kow, in order that B may vanish, we must have 

cos. = sin. ; that is, = 45°. 
Making = 45°, equation (2) becomes 

or ^*+y.2V2— af.2V2+8=0, 

which may be written 

or (7/+V2y = 2V2(x-^j. 

112 



178 ANALYTICAL GEOMETRY. 

If now we transfer the origin to a point whose co-ordinates 

are x — —r-^ and y~ — -y/2, 

the equation to the curve will become 

if = x.2^/± 
The student should construct a figure showing the form and 
position of the curve with respect to the different axes of ref- 
erence. 

234. Equation to the conic sections referred to the same 
axes and origin. When the origin of co-ordinates is placed at 
the vertex of the major axis, the equation of the ellipse (Art. 

129) is ^=jIfl«-«OS 

the equation to the hyperbola for a similar position of the or- 

igin (Art. ISO) is f = -£2ax + x*) ; 

a 

the equation to the circle (Art. 63) is 

if — 2rx—x l \ 
and the equation to the parabola (Art. S5) is 

if — iax. 
These equations may all be reduced to the form 

tf=zmx+nx\ 

2/> a -b* 

In the ellipse, m — — , and n— — 5- J 

2b* Z> a 

in the hyperbola, in — — , and n——^\ 

in the parabola, m — ia, and n = Q. 

In each case m represents the latus rectum of the curve, and 
n the square of the ratio of the semiaxes. In the ellipse n is 
negative, in the hyperbola it is positive, and in the parabola it 
is zero. 

The equation if^mx + nx 1 is the simplest form of the equa- 
tion to the conic sections taken collectively, and referred to the 
same axes and origin. 



GENERAL EQUATION OF THE SECOND DEGREE. 



179 



235. Miscellaneous Examples. 

Draw the curves of which the following are the equations : 



Ex.1, a: 5 + 2/ = 10. 

Ex.2. a5 , -V= 10 - 

Ex.3. x , + Sx=10i/. 

Ex.4. xi/+10y=4O. 

Ex.5, 3.e' + 2^ = lS. 

Ex.6. 3.*; 3 + 2/=:-lS. 

Ex.7. 3* 3 + 2^ = 0. 



Ex. 8. y»=4(«-3). 
Ex. 9. 3ajy=5. 
Ex.10. Sxy-x+2=0. 
Ex.11. 5^+7^=11. 
Ex.12. 3/-2y+4a:=0. 
Ex.13. y»+5y-9aj+10=0. 
Ex.14. 7x'-lh/=-50. 



180 



ANALYTICAL GEOMETRY. 



SECTION IX. 

TAKES OF TIIE THIRD AND HIGHER ORDERS. 

236. Lines of the third order have their equations of the form 

ay 3 + by 2 x + cyx 2 + dx 3 + ey 2 +fyx + gx 2 + liy + Jcx + 1 = 0. 
Newton has shown that all lines of the third order are com- 
prehended under some one of these four equations : 

(1) xy 2 + ey=ax 3 + bx 2 + cx + d; 

(2) xy=ax 3 + bx 2 + cx + d; 

(3) y 2 = ax 3 + bx 2 + cx + d; 

(4) y~ax 3 + bx 2 + cx+d; 

in which a, b, c, d, e may be positive, negative, or evanescent, 
excepting those cases in which the equation would thus become 
one of an inferior order of curves. 

The first equation comprehends seventy-three different spe- 
cies of curves, the second only one, the third five, and the fourth 
only one, making eighty different species of lines of the third 
order. 



237. It is not proposed to attempt any general investigation 
of the equation of the third degree, but merely to select a few 
instances calculated to exhibit the properties of some of the 
more remarkable curves. 

Ex.1. Trace the curve whose equation is (jy — x 3 . 
Suppose x — 0, then y — 0. 

a=±l, " y==fc0.167. 
a?==k2, " y=±1.333. 
a==fc3, " y=±4.500. 
x=±±, " y= ±10.667, etc. 
Constructing these values, we obtain the 
figure annexed. This equation may be 
written mure generally ay=x 3 , and the curve is called the cu- 
bical parabola. It belongs to eq. (4), Art. 230. 



LINES OF THE TIIIRD AND IIIGIIER ORDERS. 



1S1 



Ex. 2. Trace the curve whose equation is -ii/' — x 3 . 
Suppose 



a= 0, then y-- 

a?=+2, 
a=+3, 




:0. 

y=±0.500. 
y=±1.414. 

y= ±2.598. 
y= ±4.000. 
a?=+5, " y= ±5.590. 

If x is negative, ?/ becomes imaginary. The 
curve is represented by the annexed figure, and is called tire 
scmieiibieal parabola. The equation in a more general form 
is ay 2 = x 3 , and belongs to eq. (3) of Art. 236. 

Ex. 3. Trace the curve whose equation is 
t <r?/ 2 = 10. 

Suppose y = 0, then x— infinite. 

" x is negative, " y is impossible. 
" y=±l, " #=+10, etc. 

The curve is of the form represented in the 
annexed figure, and belongs to equation (1), 
Art-. 236. 



Ex.4. Trace the curve y—x 3 —x. 
Suppose x=0, then y—0. 

£=±0.5, " y= =+=0.375. 

#=±1, " y = 0. 



x=±x. 






y= ±6. 



The curve is shown in the annexed "figure, and 
belongs to eq. (4), Art. 236. 

Ex. 5. Trace the curve y 1 — x 3 —x. 
Suppose x = 0, then y—0. 

x=±l, " y—0. 

^—+0.5 " y — impossible. 

^=-0.5, " y= ±0.612. 

£=+2, u y= ±2.449. 

x=+3, " y- ±4.899. 
The curve is shown in the annexed figure. 



-e 




182 



ANALYTICAL GEOMETRY. 



/ Ex. G. Trace the curve whose equa- 

^- tlOTl is 



10y=^ 3 — 9a?" + 23aj— 15. 



Ex. 7. Trace the curve whose equa- 
tion is 

l(y=£ 3 _ll^ + 34#-24. 



Ex. 8. Trace the curve whose equa- 
tion is 

10y 2 =£ 3 -9# 5 + 24z-10. 



Ex. 9. Trace the curve whose equation is 
lCy=£ 3 -12a! 2 + 48£-04. 



Ex. 10. Trace the curve whose equation is 
10?/ 2 =^ + 3^-22^-24. 



Ex. 11. Trace the curve •whose equation is 

y=x*— 3x. 
Ex. 12. Trace the curve whose equation is 

y* = x 2 -dx. 
Ex. 13. Trace the curve whose equation is 

^=a?— of. 

238. Equations of the fourth degree. The general equation 
of the fourth degree represents an immense variety of curve 
lines, the number of different species being estimated at more 
than 5000. The number of species of lines of the fifth and 



LINES OF THE THIRD AND HIGHER ORDERS. 183 



higher orders is so great as to preclude any attempt to enumer- 
ate them completely. 



Ex. 1. Trace the curve whose equation is 

7/£ 3 = 81. 



Ex, 2. Trace the curve whose equa- 
tion is 



Ex. 3. Trace the curve whose equation is 
x y+x* + 6x*-lGx*-150x=225. 



Ex. 4. Trace the curve whose equation is 
x* + 2xy+y* = (x*+yy=16(x>-7/). 




184 



ANALYTICAL GEOMETRY. 



SECTION X. 



TRANSCENDENTAL CURVES. 



239. Equations classified. Equations may be divided into 
two classes, algebraic and transcendental. An algebraic equa- 
tion between two variables, x and y, is one which can be reduced 
to a finite number of terms involving only integral powers of 
x and y, and constant quantities. Equations which can not be 
thus reduced are called transcendental ; for they can only be 
expanded into an infinite series of terms, in which the power 
of the variable increases without limit, and the equation tran- 
scends all finite orders. 



240. Curves classified. Curves whose equations are tran- 
scendental are called transcendental curves. Among tran- 
scendental curves, the cycloid and the logarithmic curves are 
the most important. The logarithmic curve is useful in exhib- 
iting the law of the diminution of the density of the atmos- 
phere, and the cycloid in investigating the laws of the pendu- 
lum and the descent of heavy bodies toward the centre of the 
earth. 

CYCLOID. 

241. A cycloid is the curve described by a point in the cir- 
cumference of a circle rolling in a straight line on a plane. 




n 



Thus, if the circle EPN be made to roll in a given plane 
upon a straight line AC, the point P of the chmimference, 



TRANSCENDENTAL CURVES. 1S5 

which was in contact with A at the commencement of the mo- 
tion, will in a revolution of the circle describe a curve ABC, 
which is called the cycloid. The circle EPX is called the gen- 
erating circle , and P the generating point. 

When the point P has arrived at C, having described the arc 
ABC, if it continue to move on, it will describe a second arc 
similar to the first, and so on indefinitely. As, however, in each 
revolution of the generating circle an equal curve is described, 
it is only necessary to examine the curve ABC, described in 
one revolution of the generating circle. 

242. After the circle has made one revolution, every point 
of the circumference will have been in contact with AC, and 
the generating point will have arrived at C. The line AC is 
called the base of the cycloid, and is equal to the circumfer- 
ence of the generating circle. The line BD, drawn perpen- 
dicular to the base at its middle point, is called the axis of the 
cycloid, and is equal to the diameter of the generating circle. 

243. To find the equation of the cycloid. Let us assume the 
point A as the origin 
of co-ordinates, and 
let us suppose that 
the generating point 
has described the arc 
AP. If X designates A H 
the point at which the generating circle touches the base, it is 
plain that the line AN will be equal to the arc PX. Through 
N draw the diameter EX, which will be perpendicular to the 
base of the cycloid. Through P draw PII parallel to the 
base, and PR perpendicular to it. Then PR will be equal to 
IIX, which is the versed sine of the arc PX. 

Let hJX—x, and PR or IIX=?// and let r represent the ra- 
dius of the generating circle. By Geom., Bk. IV.,Prop. 23,Cor., 

RX = PII= VXlfxTlE = Vy{2r-y) = V2ry-y* ' 
also, AR=AX-RX=arc PX-PII. 




5 



18G ANALYTICAL GEOMETRY. 

The arc PN is the arc whose versed sine is UN" or y. 
Substituting the values of AR, AN, and EN, we have 
a?=(the arc whose versed sine is y)—V2ry—y\ 
which is the equation of the cycloid. 

244. Another form of the equation. It is aometimes con- 
venient, in the equation of the cycloid, to employ the angle of 
rotation of the generating circle, or the angle subtended by the 
arc PN at the centre of the circle EPN. Let this angle be 
denoted by 0, and the radius of the circle by r; then 

the arc PN=^0, 
and AH or x=zrQ— r sin. 0, 

and IXN or y=r— r cos. 0. 

If we eliminate 6 from these two equations, we shall obtain 
the same value of x as given in Art. 243. 

LOGARITHMIC CURVE. 

245. The logarithmic curve takes its name from the prop- 
erty that, when referred to rectangular axes, any abscissa is 
equal to the logarithm of the corresponding ordinate. The 
equation of the curve is therefore 

x—log. y. 

If a represent the base of a system of logarithms, we shall 
have (Alg., Art. 394) y=a x . 

To examine the course of the curve, we find, when x=0, 
y = a° = l; as x increases from to <x>, y increases from 1 to <x> ; 
as — x increases to oo, y decreases from 1 to 0. Draw AB per- 
pendicular to DC, and make it equal to the linear unit ; then 
the curve proceeding from B to the right of AB recedes from 
the axis of x, and on the left continually approaches that axis, 
which is therefore an asymptote. 

Any number of points of the curve may be determined from 
the equation y~a x . Let AC be divided into portions each 
equal to AB. Let a be taken equal to the base of the given 
system of logarithms, for example 1.6, and let a 2 , a 3 , etc., cor- 






TRANSCENDENTAL CURVES. 



187 



respond in length with the 
different powers of a. Then 
the distances from A to 1, 2, 
3, etc., will represent the loga- 
rithms of a, a?, a% etc. 

The logarithms of numbers 
less than a unit are negative, 
and these are represented by 
portions of the line AD to the 
left of the origin. 





246. In a similar manner we may construct the curve for 
any system of logarithms. Thus, for the Naperian system, 

a= 2.718. 
cf= 7.389. 
a 3 =20.085. 
a- 1 ^ 0.368. 
a- 2 = 0.135, etc. 
If at the point A we erect an 
ordinate equal to unity, at the 
point 1 an ordinate equal to 
2.718, at the point 2 an ordinate 
equal to 7.389, etc., at the point —1 an ordinate equal to 0.368, 
etc., the curve passing through the extremities of these ordi- 
nates will be the logarithmic curve for the Naperian base. 

Ex. 1. Construct by points the logarithmic curve, the base 
being 10. 

Ex. 2. Construct by points the logarithmic curve, the base 
being ^. 

CUKVE OF SINES, TANGENTS, ETC. 

247. If we conceive the circumference of a circle to be ex- 
tended out in a right line, and at each point of this line a per- 
pendicular ordinate to be erected equal to the sine of the cor- 
responding arc, the curve line drawn through the extremity of 
each of these ordinates is called the curve of sines. 



188 ANALYTICAL GEOMETRY. 

Draw a straight line ABC equal to the circumference of a 

given circle, and 
s upon it lay off the 

a m f iKr~ SG x lengths of several 

arcs, at every 10° for 
example, from 0° at 
A to 360° at C ; from these points draw perpendicular ordi- 
nates equal to the sines of the corresponding arcs, upward or 
downward, according as the sine is positive or negative in that 
part of the circle ; then draw a curve line ADBEC through the 
extremities of all these ordinates ; it will be the curve of sines. 

248. To find the equation of the curve of sines. Draw any 
ordinate PM. Let AM=^, and PM=y/ then the equation is 

If r represent the radius of the given circle, then 

y—r sin. -. 

Since the sine is when the arc is 0, the curve cuts the axis 
at A. Since the sine of 90° is a maximum, the highest point 
of the curve will be at D, where y—r. The curve cuts the 
axis again in B ; from B, y increases negatively until it equals 
— r, and then decreases to 0, so that we have a second branch 
equal and similar to the first. Beyond C the values of y recur, 
and the curve continues the same course ad infinitum. Also, 
since sin. (— x)~— sin. %, there is a similar branch to the left 
of A. 

In a similar manner may be drawn the curve of tangents, 
the curve of secants, etc. 

SPIRALS. 

249. Definition. If a right line be revolved uniformly in 
the same plane about one of its points as a centre, and if at the 
same time a second point travel along the line in accordance 
with some prescribed law, the latter point will generate a curve 
called a spiral. 



TRANSCENDENTAL CURVES. 



ISO 



Thus, let PD be a straight line which revolves uniformly 
round the point P, starting from 
the position PC, and at the same 
time let a point move from P 
along the line PD according to 
some prescribed law ; the point 
will trace out a curve line which 
commences at P, and after one 
revolution will arrive at a point 
A ; after two revolutions it will 
arrive at a point B, and so on. 
The curve thus traced is called a sjpifal. 




250. The fixed point P, about which the right line revolves, 
is called the pole of the spiral. The portion of the spiral gen- 
erated while the straight line makes one revolution is called a 
spire. If the revolutions of the radius vector are continued 
indefinitely, the generating point will describe an unlimited 
spiral. It is assumed that the point does not, after a limited 
number of revolutions, describe again the previous curve, but 
that any straight line drawn through the pole of the spiral will 
cut the curve in an infinite number of points. 

Instead of starting from the pole, the generating point may 
commence its motion at any distance from the pole; and in- 
stead of receding, it may move toward the pole. 

With P as a centre, and any convenient radius as PA, de- 
scribe the circumference ADE; 
the angular motion of the radius 
vector about the pole may be 
measured by the arcs of this cir- 
cle, estimated from A. It is gen- eI 
erally convenient to make the ra- \ 
dius of the measuring circle equal 
to the length of the radius vector 
at the end of one revolution of 




190 



ANALYTICAL GEOMETEY. 



the generating point, starting from the pole, but the measuring 
circle may have any magnitude. 

251. Spiral of Archimedes. While the line PD revolves 
uniformly round the point P, let the generating point also 
move uniformly along the line PD ; it will describe the spiral 
of Archimedes. 



252. To construct the spiral of Archimedes. Let P be the 
pole, and PX the first position of the 
radius vector. "With P as a centre, 
and any convenient radius, describe 
the measuring circle ACEGr, and di- 
vide its circumference into any con- 
venient number of equal parts, as, for 
example, eight. On PB set off PI any convenient distance ; 
on PC set off PK=2PI; on PD set off PL=3PI, etc. The 
curve passing through the points I, K, L, M, etc., thus deter- 
mined, will be the spiral of Archimedes, for the radii vectores 
are proportional to the arcs AB, AC, etc, of the measuring 
circle. 




253. To find the equation to the spiral of Archimedes.- From 
the definition of the curve, the radii vectores and the measur- 
ing arcs increase uniformly ; that is, in the same ratio. Hence 
we have 

PL : PR : : angle APD : four right angles. 
Designate the radius vector PL by r, PR by &, and the variable 
angle by ; then we shall have 

r:$::0:2ir; 

bO . h 

whence P=q~ ; or, putting 0=5-, we have the equation 

r=a0. 
When the radius vector has made two revolutions, or — ^tt^ 
we have r—2b; that is, the curve cuts the axis PX at a dis- 
tance equal to 2PR; after three revolutions it cuts the axis 



TRANSCENDENTAL CURVES. 



191 



PX at a distance equal to 3PR, etc. Hence the distance be- 
tween any two consecutive spires, measured on a radius vector, 
is always the same. 

254. Hyperbolic spiral. While the line PN revolves uni- 
formly about P, let the generating point move along the line 
PjST in such a manner that the radius vector shall be inversely 
proportional to the corresponding angle ; it will describe the 
hyperbolic spiral. 

255. To find the equation to the hyperbolic spiral. From 
the definition of the curve, the 
radius vector is inversely propor- 
tional to the measuring angle; 
hence we have 

PG : PN : : angle APN : four 
right angles. 
Designate the radius vector PN 
by r, PG by 5, and the variable 
angle measured from the line 
PX by 0, and we shall have 

b:r::0: 

Whence r0 = 2b7r; 

or, putting 2bir = a, we have 

rQ — a. 
When 0=0, r=oo; as 6 increases, r decreases, at first very 
rapidly, but afterwards more uniformly. As may increase 
without limit, r may decrease indefinitely without actually be- 
coming zero ; hence, as the radius vector revolves, the curve 
continues to approach the pole, but reaches it only after an in- 
finite number of revolutions. This curve is called the hyper- 
bolic spiral from the similarity of its equation to that of the 
hyperbola referred to its asymptotes (xy—c' 2 ), the product of 
the variables r and being equal to a constant quantity. 




2tt. 



192 



ANALYTICAL GEOMETRY. 



256. To construct the hyperbolic spiral. Let P be tlie pole, 
and PX the first position of the radius vector. With any con- 
venient radius draw the measuring circle ABDE, and divide 
its circumference into any convenient number of equal parts 
AB, BO, CD, etc. On PB, produced if necessary, take any con- 
venient distance, as P1ST ; take PM equal to one half of PN, 
PL equal to one third of PN", PK equal to one fourth of PjST, 
etc. ; the curve passing through the points N", M, L, K, etc., will 
be an hyperbolic spiral, because the radii vectores are inverse- 
ly proportional to the corresponding angles measured from 
PX. 



257. Logarithmic spiral. While the line PA revolves uni- 
formly about P, let the generating point move along PA in 
such a manner that the variable angle may be proportional to 
the logarithm of the radius vector; it will describe the loga- 
rithmic spiral. 

The equation of the logarithmic spiral is 



= lO; 



&• 



a' 



or 



r = ab% 

I) being the base of the system of logarithms (Alg., Art. 394), 
and a any arbitrary constant. 



258. To construct the logarithmic spiral. If we take Z> = 10, 

the base of the common system of 
logarithms, the changes of r are so 
rapid that we can represent only a 
small arc of the curve. We will 
therefore assume 5=1.2. When 
= 0, r = a, which determines the 
point L. When = 1, that is, 57°.3 
(radiwbeing unity), 7 , =1.2^,which 
determines the point M. When 
= 2, that is, 114°.G, r = 1.2 2 a, or 1.44a, which determines the 




TRANSCENDENTAL CURVES. 193 

point N, etc. As increases, r also increases, but does not be- 
come infinite until 6 becomes infinite. 

If we suppose the radius vector to revolve in the negative 
direction from PA, when 0=— 1,^ = 0.83(2, which determines 
another point of the curve. "When 0=— 2, r — 0,6da, etc. 
Hence we see that, as the radius vector revolves in the nega- 
tive direction, it generates a portion of the spiral which slowly 
approaches the pole, but can not reach it until 0=— -oo. 

Thus we see that the logarithmic spiral makes an infinite 
series of convolutions around the pole P. 

I 



194 ANALYTICAL GEOMETRY. 



PAET III 

GEOMETRY OF THREE DIMENSIONS. 

SECTION I. 
OF POINTS IN SPACE. 

259. Hitherto we have considered only points and lines 
uated in one plane, and we have seen that the position of a 
point in a plane may be denoted by its distances from two as- 
sumed fixed lines or axes situated in that plane. We have now 
to consider how the position of any point in space may be rep- 
resented. 

260. To determine the positio?i of a point in space. Let 
three planes XAY, ZAX, ZAY, supposed to be of indefinite ex- 
tent, be drawn perpendicular to each other, and let these planes 
intersect each other in the three straight lines AX, AY, AZ. 
Let P be any point in space whose position it is required to 
determine. 

From the point P draw the line PB perpendicular to the 
plane XAY ; draw PC perpendicular to 
the plane ZAX, and PD perpendicular to 
the plane ZAY; then the position of the 
point P is completely determined when 
__ x these three perpendiculars are known. 

Let a, b, c represent these three perpen- 



/% b diculars. On AX take AE = a, on AY 

Y take AF = £, and on AZ take AG = c, and 

through the points E, F, and G let planes be drawn parallel to 
the three planes ZAY, ZAX, and XAY, forming the rectangu- 
lar parallelopiped EFG. 

Since the plane drawn through E is every where distant from 
the plane ZAY by a quantity equal to a, the point P must be 



OF POINTS IN SPACE. 195 

somewhere in this plane ; and since the plane drawn through 
F is every w T here distant from the plane ZAX by a quantity 
equal to b, the point P must be also in this plane. It must 
therefore be in the line BP, which is the common section of 
these two planes. Also, since the plane drawn through G is 
every where distant from the plane XAY by a quantity equal 
to c, the point P must be somewhere in this plane ; it must 
therefore be at the intersection of this third plane with the 
line BP. Thus the position of the point P is completely de- 
termined. 

261. Definitions. The three planes XAY, ZAX, ZAY, by 

reference to which the position of the point P has been deter- 
mined, are called the co-ordinate planes. The first is desig- 
nated as the plane XY, the second as the plane XZ, and the 
third as the plane YZ. The lines AX, AY, AZ, which are the 
intersections of these three planes, are called the co-ordinate 
axes. The first is called the axis of X, and distances parallel 
to it are denoted by x ; the second is the axis of Y, and dis- 
tances parallel to it are denoted by y ; the third is the axis of 
Z, and distances parallel to it are denoted by z. The point A, 
in which the three axes intersect, is called the origin of co-or- 
dinates. The equations of a point in space are therefore of 
the form x=a, y=b, z—c. 

262. Signs of the co-ordinates. If the three co-ordinate 
planes be indefinitely produced, there will 
be formed about the point A eight solid an- 
gles, four above the horizontal plane XAY, 
and four below it. It is required to denote x 
analytically in which of these angles the 
proposed point is situated. For this pur- 
pose, if we regard distances measured on 
AX to the right of A as positive^ we must regard distances 
measured to the left of A as negative. So, also, y is regarded 
as positive when it is in front of the plane ZX, and negative 

m 





it 


it 


a 


ZXAY'. 


u 


it 


a 


ZX'AY'. 


it 


a 


a 


ZX'AY. 


u 


a 


a 


ZXAY. 


it 


a 


a 


ZXAY'. 


it 


a 


a 


ZXAY'. 


it 


a 


it 


ZX'AY. 



196 ANxVLYTICAL GEOMETRY. 

when it is behind that plane; and z is regarded as positive 
when it is above the plane XY, and negative when it is below 
that plane. Hence the equations of a point in each of these 
eight angles are as follows : 
If x= + a, y= +b, z= +c, the point is in the angle ZXAY. 

x=-\-a, y=—b,z=+c, 

x=—a, y=—b, 0= + <?, 

x=—a } y= +b y z=+c, 

x=+a, y=+b,z=~c y 

x=+a, y=—b, z——c, 

x— — #, y=—b, z = —c, 

x=—a, y= +b, z= — <?, 

263. Co-ordinates of particular points. If the point P be 
situated in the plane of xy, then its distance z from this plane 
is 0, and its equations will be 

x=±a, y=zkb, z = 0. 
If the point be situated in the plane of xz, then its distance 
y from this plane is 0, and its equations will be 
x==a, y = 0, z==c. 
If the point be situated in the plane of yz, then its distance 
x from this plane is 0, and its equations will be 
x = 0, y=±b, z==c. 
If the point be situated on the axis of x, that is, on the inter- 
section of the planes xy and xz, then its distance from each of 
these planes is 0, and its position will be expressed by the equa- 
tions x=±a, y — 0, z=0. 
So, also, if the point be situated on the axis of y } we shall have 

x = 0, y=±b, z = 0; 
and if it be situated on the axis of z, we shall have 
x = 0, y = 0, z==c. 
If the point be at the origin, its position will be denoted by 
the equations cc = 0, y=0, z = 0. 

Ex.1. Indicate by a figure the position of the point whose 
equations arc 

x=+±, y=-3, s=-2. 






OF POINTS IN SPACE. 



197 



Ex. 2. Indicate by a figure the position of tlie point whose 



equations are 



x—- 



y=+7, z=+5. 



Ex. 3. Draw a triangle, the co-ordinates of whose angular 
points are x— + 3, y — + 4, z — + 2 ; 

a>=-3, y=-4, s=-2; 
#=— 1, y-0, s=+l. 

264. Projections. If a perpendicular be let fall from any 
point P upon a given plane, the point in which this line meets 
the plane is called the projection of the point P on the plane. 
The projections of the point P (Art. 260) on the three co-ordi- 
nate planes are the points B, C, D. 

The projection of any curve upon a given plane is the curve 
formed by projecting all of its points upon that plane. When 
the curve projected is a straight line, its projection on any one 
of the co-ordinate planes will also be a straight line, for all the 
points of the given line are comprised in the plane passing 
through this line and drawn perpendicular to the co-ordinate 
plane ; and since the common section of any two planes is a 
straight line, the projections of the points must all lie in one 
straight line. This plane, which contains all the perpendicu- 
lars drawn from different points of the straight line, is called 
the projecting plane. 

If the positions of any two projections of the point P are 
given, it will be sufficient to determine the point P ; for a line 
drawn from either projection, perpendicular to the plane in 
which it is, necessarily passes through the point P, so that P 
will be at the intersection of two such perpendiculars. When 
two projections of a point are known, we can always determine 
the third. 



265. To find the distance of any point 
from the origin in terms of the co-ordi- 
nates of that point Let AX, AY, AZ be 
the rectangular axes, and P the given point. 
Let the co-ordinates of P be AE=#, BE=y, 
andPB=3. 




198 , ANALYTICAL GEOMETRY. 

The square on AP = the sum of the squares on AB and PB. 

Also, the square on AB = the sum of the squares on AE 
and EB ; that is, AP 2 = AE 2 + EB 2 + PB 2 , 
or AT 2 =:x'+f + z\ 

Ex. 1. Determine the distance from the origin to the point 
whose co-ordinates are 

x—2a, y=—3a, z — Qa. 

Ex. 2. Determine the distance from the origin to the point 
whose co-ordinates are 

x——b, y——^b, z — Sb. 

266. To find the distance between two given points in space. 
Let M and N be the two given points, their co- 
-$ ordinates being respectively x, y> z, and x\ y f , z' . 
n If the points M and N be projected on the 
x plane of xy, the co-ordinates x, y of the projec- 
tions m and n will be the same as those of the 
points M and N. Hence, for the distance mn 
we have (Art. 21) 

mn q =(x— x r y + (y— y') 2 . 
Now, if MB be drawn parallel to mn, MRN will be a right 
angle, and hence MN 3 =MR 2 + NR 2 

=MR 2 + (N^-Rn) 2 ; 
that is, MN = V(x-x'y + (y-yy + (z-zy ; 

that is, the distance between any two given points is the diag- 
onal of a right parallelepiped, whose three adjacent edges are 
the differences of the parallel co-ordinates. 

Ex.1. Determine the distance between the points 
x~3, 2/=4, and z~—2, 
and x=4:, ?/=— 3, and z — \. Ans. V^- 

Ex. 2. Determine the distance between the points 
x=2, y = 2, = 1, 
and x—— 2, y=— 3, s = 4. Ans. 






THE STRAIGHT LINE IN SPACE. 



199 



X 



SECTION II. 

THE STRAIGHT LINE IN SPACE. 

267. A straight line may be regarded as the common section 
of two planes, and therefore its position will be known when 
the position of these planes is known ; hence its position may 
be determined by the projecting planes, and the situation of 
the projecting planes is given by their intersections with the 
co-ordinate planes; that is, by the projections of the given line 
upon the co-ordinate planes. 

268. To find the equation of a straight line in space. 
Let x=mz+a 

be the equation of a straight line Mj/ 
in the plane of xz, and through this line 
let a plane be drawn perpendicular to 
the plane xz. Also, let 
y — nz+b 
be the equation of a line mp in the 
plane of yz, and through this line let a / x 
plane be drawn perpendicular to the plane yz. These two 
planes will intersect in a line MP, which will thus be com- 
pletely determined. The two equations 

x — mz+a^ (1) 

y=nz + b, (2) 

taken together, may therefore be regarded as the equations of 
the line MP, and from these equations the line MP may be 
constructed ; for, if a particular value be assigned to either va- 
riable in these equations, the values of the other two variables 
can be found, and these three quantities taken together will be 
the co-ordinates of a point of the required line. 

Thus, suppose n'r to be a value of z ; this, with the corre- 
sponding value of x deduced from equation (1), will determine 





z 
















sM 


?i' 




71 


1/ 










1 












/ 


A 




r 


/ P/ 


/ 




3 


R\ 




V 









200 ANALYTICAL GEOMETRY. 

a point ?i\ through which a line must be drawn perpendicular 
to the plane xz. The same value of £, with the corresponding 
value of y deduced from equation (2), will determine a point 
n, through which if Nn be drawn perpendicular to the plane 
yz, it will intersect the line ISm', since both lines are situated in 
the same plane, viz., a plane parallel to xy, and at a distance 
from it equal to z. The point N of the line MP is therefore 
determined, and in the same manner we may determine any 
number of points of this line. Hence the equations to the 
straight line MP are x = mz + a 9 (1) 

yz=znz+b. (2) 

289. Interpretation of the constants in these equations. In 
equation (1) m represents the tangent of the angle which the 
projection of the given line on the plane xz makes with the 
axis of z, and a represents the distance cut from the axis of X 
by the same projection (Art. 29). 

In equation (2) n represents the tangent of the angle which 
the projection on the plane yz makes with the axis of £, and b 
is the distance cut from the axis of Y. 

If we combine these two equations, and eliminate the varia- 
ble £, we shall have n 

y—b = —(x—a), 

which expresses the relation between the co-ordinates of the 
point R, which is the projection of the point N" on the plane 
xy, and therefore this last equation is the equation of the line 
MP projected on the plane xy. 

Ex. The equations of the projections of a straight line on the 
co-ordinate planes zx } zy are 

#=23 + 3, y=3s— 5; 

required its equation on the plane xy. Ans. 2?/ = 3 t £— 19. 

270. To determine the points where the co-ordinate planes 
are pierced by a given straight line. At the point where a 
line pierces the plane xy the value of z must be 0. If we sub- 






TIIE STRAIGHT LINE IN SPACE. 201 

stitute this value of z in equations (1) and (2) of Art. 268, we 
shall find x = a, y = b ; 

hence a and b taken together are the co-ordinates of the point 
in which the given line pierces the plane xy. 

In like manner, the co-ordinates of the point in which the 
line pierces the plane xz may be determined by putting y—0 
in equation (2), and substituting the resulting expression for z 
in equation (1). In the same manner, the point where the line 
pierces the plane yz may be determined. 

Ex.1. Determine the points where the co-ordinate planes are 
pierced by the line whose equations are 

# = 2,2 + 3, 
y=3z-7. 
Ex. 2. Determine the points where the co-ordinate planes are 
pierced by the line whose equations are 

x—— 2z—5 y 
y=-z+2. 

271. To find the equations of a straight line passing through 
a given point. Let the co-ordinates of the given point be x', 
y', z\ and let the equations to the straight line be 

x = m2 + a, y=nz+b. 
Kow, since this line passes through the given point, we must 



have 


x' = ?nz' + a> 




y'—nz'+b; 


hence we obtain 





x—x'=m(z—z f ), 
and y— y'=n(s— a*), 

which are the equations sought, and characterize every straight 
line which can be drawn through the point x\ ?/', z\ If the 
given point be the origin, then #' = 0, y f — 0, and s'=0 3 and the 
equations of a line passing through the origin are 

x—7nz y y—nz. 

272. Equations of a straight line passing through two given 
points. Let the co-ordinates of the given points be x\ y\ z\ 

12 



202 ANALYTICAL GEOMETEY. 

and x\ y'\ z" ; then the equations of the line passing through 
the first of these points are 

x-x f = m{z-z%\ 

y-y'=n{z-z f ).\ W 

Since the line passes through the point x h ', y", z f \ we must also 
have x" —x f — m(z f/ — z'), 

and y"-y'=n(z"-z% 

from which we obtain the values of m and n, viz. 

x' f —x r y"—y' 



Z" — Z'? z"—z' 

These values of m and n, being substituted in equation (1 
will famish the equations of the line passing through both the 
given points. We have, therefore, 

/ & 3? f 

X — X — z ff_ z \2 — 2), 

If one of the points a?", y"> z" be the origin, these equations 

become x— — . s, 

z 

y' 

Ex. 1. Find the equations to the straight line passing through 
the following points : 

a?'=3, y'=-4j s'=2, 

x"=-$, y"=6, s"=3. 

Ans. x= —83 + 19, y=10z-2±. 
Ex. 2. Find the equations to the straight line passing through 
the following points : 

aj'=4, ;?/ = -2, 3'=-3, 
x" = 0, y"=l, z"=—% 

Aiis. x=—4:Z—8, y=3z+7. 

273. To determine the conditions requisite for the intersec- 
tion of two straight lines. Two straight lines which are not 
parallel must meet if they are situated in the same plane, but 



THE STRAIGHT LINE IN SPACE. 203 

tliis is not necessarily true for lines situated any where in space. 
In order that two lines may meet, there must be a particular 
relation among the constant quantities in their equations. In 
order to discover this relation, let the equations to the lines be 

x=?nz + a, \ x — ori r z-\-a' ', ) 

y=nz + b, J and y=n'a+V. ) 
If these lines intersect, that is, have one point in common, the 
co-ordinates of this point must satisfy both sets of equations, 
or for this point the values of x, y, and z must be the same in 
all the equations. Since x of the one line equals x of the oth- 
er, we have (m — m')z + a— a/ = 0, 

a' — a 

or z — ,; 

m—m' 

and since y of one line equals y of the other, we have 

(n-n')z + b-U = 0, 

V-b 

or z— -,. 

n—n 

But z of the one line is equal to z of the other ; hence 

a' -a V-b 
in—m'~~n—n f ' 
Hence, when the lines intersect, the relation between the con- 
stants is given by the equation 

(a f -a)(n-n') = {U-b)(m-m f ). (1) 

Conversely, when this equation exists the two lines intersect. 

The co-ordinates of the point of intersection may be deter- 
mined by substituting in the expressions for x and y the value 
of z just found. They are 

ma!—m!a nb'—n'b 

x= r~i V— j-* 

m—m ' * n—n 7 

_ a'-g V-b 
~ 'm—m'~ n—n' % 
These values of x and y, with either value of 2, will give a 
point of intersection w T hen equation (1) is satisfied. 

If m = m\ and n = n\ equation (1) is satisfied, and the values 
of x, y, and z become infinite. The point of intersection is 
then at an infinite distance; that is, the two lines are parallel. 



201 ANALYTICAL GEOMETRY. 

But when m=m / , the projections of the two lines on the 
plane xz are parallel, and when n — n r the projections on the 
plane yz are parallel. Hence, iftivo right lines in space are 
parallel^ their projections on the same co-ordinate plane will 
he parallel. 

274. To find the equations of the straight line which passes 
through a given point and is parallel to a given line. Let 
x\ y\ z' be the co-ordinates of the given point. The equations 
of the straight line passing through this point (Art. 271) are 

x— x' = ?n(z— z f ), 
and y—y'=n(z—z r ). 

In order that this line may be parallel to a given line, its 
projections on the co-ordinate planes must be parallel to the 
projections of the former line (Art. 273) ; that is, they must 
cut the axis of z at the same angle. The quantities m and n 
therefore become known, and if we represent the tangents of 
the given angles by mf and n\ we shall have 

x—x'=m/(z—z% 
y-y'=:n'(2--z'), ^ 
which are the equations of the required line. 

Ex. Find the equation of a straight line which passes through 
the point x f = 3, y f = — 2, z r = 1, 

and is parallel to the line whose equations are 
x = <iz + 5, y=—z + 3. 

275. To find the relation which exists among the angles 
which any straight line makes with the axes of co-ordinates. 
Let a, j3, and y represent the angles which the straight line 

makes with the axes of x, y, and z. From 
the origin, draw a line AP parallel to the 
proposed line ; the angles which it makes 
c with the co-ordinate axes will be the same 
as those made by the proposed line. In 
AP take any point P, and from it draw a 
line perpendicular to each of the co-ordinate planes. In the 



L. 



THE STRAIGHT LINE IN SPACE. 205 

triangle APG, right-angled at G, we have AG = AP cos. 7; 
also, in the triangle APF, right-angled at F, we have AF — AP 
cos. ]3 ; and in the triangle APE, right-angled at E, we have 
AE=AP cos. a. But by Art. 265 we have 
AE 2 +AF 2 +AG 2 =AP 2 ; 
hence AP 2 cos. 2 a + AP 2 cos. 2 j3 + AP 2 cos. 2 7 = AP 2 ; 
or, dividing by AP 2 , we have 

cos. 2 a + cos. 2 /3 + cos. 2 y = 1 ; (1) 

that is, the sum of the squares of the cosines of the angles 
which any straight line maizes with the co-ordinate axes is 
equal to unity. 

If it is required to determine the value of each cosine, let 

x = mz, y—nz^ 

be the equations of the line AP (Art. 271). Then 

cos. a — Tn cos. 7, and cos. fi = n cos. 7. 

Substituting these values in equation (1), we obtain 

m 2 cos. 2 7 + ri* cos. 2 7 + cos. 2 7 = 1, 

whence cos. 7: 



Vni 


> + n 
on 


•+i' 


~~Vm' 


l + n' 


'+1' 




n 





also, cos. a 

and cos. i3 — 

H Vm 2 +n*±l 

In these equations, m denotes the tangent of the angle which 

the projection of the proposed line upon the plane xz makes 

with the axis of z; and n denotes the tangent of the angle 

which the projection on the plane yz makes with the axis of z. 



206 



ANALYTICAL GEOMETKY. 



SECTION III. 

OF THE PLANE IN SPA^E. 

276. The equation of a surface is an equation which ex- 
presses the relation between the co-ordinates of every point of 
the surface. 

Three points, not in the same straight line, are sufficient to 
determine the position of a plane (Geom., Bk. VII., Prop. 2, 
Cor. 1) ; hence, if we know the points where a plane BCD in- 
tersects the three co-ordinate axes, the po- 
sition of the plane will be determined. 

The intersections of any plane with the 
co-ordinate planes are called its traces. 
Thus BC is the trace of the plane BCD 
on the plane XY, BD is its trace on the 
plane ZX, and CD is its trace on the plane 
ZY. 




277. To find the equation to a plane. Let AX, AY, AZ be 
three rectangular axes, and let BCD be the plane whose equa- 
tion is required to be determined. Let the plane intersect the 
axes in the points B, C, D, and let AB be denoted by a, AC by 
&, and AD by c. Take any point P in the given plane, and 

through P draw the plane EGII 
parallel to the co-ordinate plane 
YZ, and cutting the given plane 
and the other co-ordinate planes 
in the triangle EGH. Draw PR 
perpendicular to the plane YX. 
Then will the co-ordinates of the 
point P be 
x= AE, 2/=ER, and s=PE. 
It is required to find an equation between these co-ordinates 
and the intercepts a, &, and c. 




OF THE PLANE IN SPACE. 207 

By similar triangles BAC, BEG, we have 
BA:AC::BE:EG, 
or a:b::a— #:EG. 

OX 

Hence EQt=6-—: 

a' 

J)X 

also, KG = £— y— — > 

Again, by similar triangles D AC, PEG, we have 
DA:AC::PK:KG, 

or c:o::z:o—y— — / 

whence abz—abc—acy— hex, 

or icx+acy+abz=zabc, (1) 

or -+!+ ! =l, (2) 

which is the equation of a plane in terms of its intercepts on 
the three axes. This equation is similar in form to the equa- 
tion of a straight line (Art. 42). If we represent the coefficients 
of x, y, and z in eq. (1) by A, B, and C, this equation assumes 
the form Ax + By + Cz + D = 0, (3) 

being a complete equation of the first degree containing three 
variables, and this is the form in which the equation of a plane 
is usually written. 

278. Having given the equation ofajplane, to determine the 
equations of its traces. Let the equation of the plane be 

A#+By+Cte+D=0; 

then, for every point in this plane which is situated likewise in 
the plane of xy, that is, for every point in the trace on the plane 
of xy, we must have s=0. Hence the equation of this trace is 

Ax+By+T> = 0. (1) 

In like manner, for every point in the trace on the plane of 
xz, we must have y=0 ; hence the equation of this trace is 

Ax+Cz+T> = 0. (2) 

So also the equation of the trace on the plane of yz is 
By+Cz+D = 0. (3) 



208 ANALYTICAL GEOMETRY. 

If in equation (1) we make ?/ = 0, the resulting value of x, 

viz., — -r-, will be the distance from the origin to the point where 

the given plane meets the axis of x. If we make x=0, we 

have y=— =d for the distance from the origin to the point 

where the plane meets the axis of y. If in equation (2) we 

make x = 0,we have z~ — p for the distance from the origin 

to the point where the plane meets the axis of z. 

If D = 0, the proposed plane must pass through the origin. 
Ex. 1. Find the traces of the plane whose equation is 

2x— Sy+Tz-r 10=0. 
Ex. 2. Determine where the plane whose equation is 
3x+fy + 5z— 60 = 
meets the three co-ordinate axes. 

Ans. #=20, y=15, 3=12. 
Ex. 3. Determine where the plane whose equation is 
3£-.4?/+22+12=:0 
meets the three co-ordinate axes. 

279. To find the equation of the plane which passes through 
three given points. If in equation (2), Art. 277, we represent 
the coefficients of x, y, and z by M, N, and P, the equation of 
the plane will become 

Mx+Xy+Tz=l. (1) 

Let the co-ordinates of the three given points be 
x',y',z'; x",y",z"; x'",y'",z[". 

Since the plane passes through the three given points, the 
co-ordinates of each of these points must satisfy the equation 
of the plane, so that we must have 

M#'+]Ny+P3'=i, 

Ms"+Ny"+P3"=l, 

M£ /// +Jsy // +P3 /// =i. 

From these three equations the values of the three constants 
M, N, and P may be determined, and if these values are sub- 



OF THE PLANE IN SPACE. 209 

Btituted in equation (1), we shall have the equation of a plane 
passing through the three given points. 

Ex. 1. Find the equation of the plane passing through the 
three paints 

x'=l, */=-%, s'=_3. 

®"=2, y"=l, s"=0, 
«'"=_2, y'"=2, 0"'=-l. 

^725. 6aj + lly-133-23 = 0. 
Ex. 2. Find the equation of the plane passing through the 
three points x' = 3, y'=%, £'=4, 

®"=0, y"=4, 0"=1, 
35'"= -2, y"'=l, ^"sO. 

J.n*. lla>-3y-133+25 = 0. 

280. T<9 determine the conditions which must subsist in 
order that a straight line may be parallel to a plane. Let the 
equations of the straight line be 

x=mz+a, y—nz-\-b y 
and let the equation of the plane be 

Aaj+By+Cte+D=0. 
If through the origin we draw a straight line parallel to the 
given line, its equations will be 

x=mz, y—nz; 
and if through the origin we also draw a plane parallel to the 
given plane, its equation (Art. 278) will be 

A#+B?/+Cfe:=0. 
Now, if the first line be parallel to the first plane, the line 
drawn through the origin must coincide with the plane drawn 
through the origin; hence the co-ordinates x and y of this 
straight line must satisfy the equation of the plane. If we 
substitute the values of x and y in the equation of the plane, 
we have Amz + B?iz + Cz = ; 

or, dividing by z, we have 

Am+Bn+G=0 3 
which is the analytical condition that a right line shall be par- 
allel to a plane. 



210 ANALYTICAL GECXMETKY. 

281. To determine the conditions which mast subsist in 
order that two planes may be parallel. Let the equations of 
the two planes be Ax+By + Cz + D =0, 

A'x+B'y+C'z+iy=0. 

The traces of these planes on either of the co-ordinate planes 

must be parallel, otherwise the two planes would meet. The 

equations of the traces on the plane of xz (Art. 2 78) are 

AD A/ W 

Z — pi£ p ? Z — ^,X — p/. 

If these traces are parallel, we must have 

A_A^ 

o~C" 

Comparing the traces on the other co-ordinate planes, we shall 

. B B' A A' 

also nna -^—-^^ ^=g>. 

The last equation could be derived from the two others, and 
hence the three equations express but two independent condi- 
tions. 

282. If a straight line be perpendicular to a plane ^ the pro- 
jection of this line on either of the co-ordinate planes toill be 
perpendicular to the trace of the given plane on that co-ordi- 
nate plane. 

Let MX be the co-ordinate plane, ABCD the proposed plane, 

EH the line perpendicular to it, and 
let GH be the projection of EH on 
the plane MX. The projecting plane 
EGH is perpendicular to MX ; and 
since the line EH is in the plane 
~a~" EGH, the plane EGII is perpendic- 

ular to the plane BD (Geom., Bk.VIL, Prop. 6). Hence the 
plane EGII is perpendicular to each of the planes MX and BD ; 
it is therefore perpendicular to their common section AB 
(Geom., Bk.VIL, Prop. 8). Hence AB, which is the trace of 
the given plane on the plane MX, is perpendicular to the plane 
EGII, and is therefore perpendicular to the line GH, which it 




OF THE PLANE IN SPACE. 211 

meets in that plane (Geom., Bk.VIL, Def . 1) ; that is, GH, which 
is the projection of the given line EH, is perpendicular to AB, 
which is the trace of BD on the plane MN". 

283. To determine the conditions which must subsist in 
order that a straight line may be perpendicular to a plane. 

Let the equation of the plane be 

A£ + B?/ + C2 + D:=:0, 

and let the equations of the projections of the straight line be 

x=?nz+a, y=nz+b. 
The equation of the trace of the plane on xz is 

Az + Cs+D^O, 

C D 
or B== __3__ : 

The equation of the trace on yz is 

By+Cz+D = 0, 
C D 

But since the projections of the line must be perpendicular to 
the traces of the plane (Art, 282), we shall have (Art. 46) 

A B 

m = p, and n — ^ 

which are the conditions required. 

284. To find the equation of a plane that passes through a 
given pointy and is perpendicular to a given straight line. 

Let x\ y \ z' be the co-ordinates of the given point, and let 
the equations to the given line be 

x — mz + a^ and y = nz+h. 
Also, let the equation of the plane be 

Ax+By+Cz+D=zO. 
Since the point (x\ y\ z') is in this plane, we have 

Aaj , + By' + C < s / + D = 0; 
hence A(x-x') + B(y— y') + C(z—2') = 0, 

which is the equation of any line passing through the point 
(x\ y f , z'). But by Art. 283 we have 



212 ANALYTICAL GEOMETEY. 

A = mC, and ~B=nC; 
hence mC(x—x') + nC(y—y') + C(2—z') = 0, 

or m(x— x') + ?i(y— y') + {z — z') = 0> 

which is the equation required. 

285. To find the equation of a straight line drawn from the 
origin perpendicular to a given plane, and determine its length. 
Let the equation of the given plane be 

A^+B?/+C2+D = 0. (1) 

The equations of a line passing through the origin are 

x—mz, y~nz. 
But if this line be perpendicular to the plane, we must have 

A B 

(Art. 283) m = ™ and n = p ; 

hence the equations of the line passing through the origin and 
perpendicular to the plane are 

As Bz 

Those values of x, y, and z, which, when taken together, will 
satisfy equations (1) and (2) at the same time, must be the co- 
ordinates of a point common to the line and plane ; therefore, 
by combining these equations, and deducing the values of x, y, 
and z, we shall obtain the co-ordinates of the point in which 
the line pierces the plane. The distance of this point from the 
origin may then be found by Art. 265. 

If P represent the length of the perpendicular, we shall have 

p= d 



VA , +B 9 +C' 

Ex. 1. Find the equations of a straight line passing through 
the origin and perpendicular to the plane whose equation is 

2x-4y+z—8 = 0. 
Find, also, the point in which the line pierces the plane, and 
find the length of the perpendicular. 

Ans. The equations of the line are x — 2z, y——^z; 

, i • . 16 32 8 

it pierces the plane in the point #=q?> V— — ^rr> 2= 2i> 

and the length of the perpendicular is 



Viil 



OF TIIE TLA.NE IN SrACE. 213 

Ex. 2. Find the length of the perpendicular from the origin 
upon the plane whose equation is -. 

2aM-3y+4s-12=0. Ans.-y=. 

286. To find the equations of the intersection of two given 
planes. 

Let the equations of the two planes be 
AaH-By+C3 + D=0, 

A'»+B'y+C's+D'=a 

If the given planes intersect, the co-ordinates of their line of 
intersection will satisfy at the same time the equations of both 
planes. If, therefore, we combine the two equations and elim- 
inate z, we shall obtain an equation between x and y, which is 
the equation of the projection on the plane xy of the intersec- 
tion of the planes. 

In a similar manner we may find the equation of the pro- 
jection of the intersection on the plane xz. But the equations 
to the projections of a line on two co-ordinate planes are the 
equations to the line itself ; hence the two equations thus found 
are the required equations to the intersection. 

Ex. Find the equations to the intersection of two planes of 
which the equations are 

2x + 5y-3z + 6 = Q, 
3x+ y + z +4 = 0. 

■ |Ute + 8z +14=0, 



214 ANALYTICAL GEOMETRY. 



SECTION IV. 

OF BUR] : r REVOLUTION. 

287. Definitions. A wZ/tf <>/" revolution is a solid which 

may bo generated by the revolution of a plane surface about a 

fixed axis. 

A surface of revolution is a surface which may be generated 
by the revolution of a line about a fixed a 

The revolving line is called the generatrix^ and the line about 
which it revolves is called the axis of the surface or solid, or 
the axis of revolution. The section made by a plane passing 
through the called a meridian section. 

It follows from the definition thai "ion made by a 

plane perpendicular to the fixed as circle whose centre 

in that a 

238. The number of solid-; of revolution is unlimited, but 
the !j are of most frequent use are the cylinder^ con 

sphere^ spJieroid,jwraboloid) and hyperboloid. 

jatian to a surface of revolution i i \t when I 

I revolution coincides with one of the co-ordinate 
In the : Qg prob! ill suppose the axis of revolu- 

ith the axis of 2, and the co-ordinate plan* 
to be ber, 

289. To find the equation to the 
of a right cylinder. A right cylinder m 
be d to be generated by the revolu- 

tion of an ; about one of . as 

/ 1/ ■ ( E of a re( ■ 

it i .t the oppo 

plain that any point of CE, as I> ; 

in it '■': of a circl 



N M 




OF SURFACES OF FwEVoLUTIOX. 



215 



Let AX, AY, AZ be the rectangular axes to which the cylin- 
der is referred, having the origin at the centre of the base of 
the cylinder, and let the axis of z coincide with the axis of the 
cylinder. 

Let the co-ordinates of any point P on the surface be AH"=3, 
XM = ;/\ and MP=yy then the square on XP = the sum of the 
squares on XM and MP, or 

FS* = x' + y\ 
But PX, which equals DX, is a constant quantity, and z may 
have any value whatever, so that the equation of a right cylin- 
der is x x +y' x — c X j z being late. 

290. To find the equation to tht :-e of a right a 

A right cone may be supposed to be generated by the revolu- 
tion of a right-angled triangle about one of its perpendicular 
sides as an axis, the hypothenuse generating the curved surface, 
and the remaining perpendicular side generating the base. 

Let AC be the hypothenuse of a right-angled triangle, an 1 
let it be revolved about AB as an axis ; 
then any point of AC, as D, in its revolu- 
tion will describe the circumference of a 
circle. 

Let the origin be placed at the vertex of 
the cone, and let the axis of z coincide with 
the axis of the cone ; then, as in Art. i 
we shall have PX' = re* + ?f. 

Let v represent the angle BAC, or the 
semiangle of the cone ; then 

XP = XD = AX tang. CAB=AN 
that is, x^+^ — z 2 tang. V, 

which is the equation of the surface of a right cone. 

If the generatrix AC is of indefinite length, the whole sur- 
face generated consists of two symmetrical portions, each of 
indefinite extent, lying on opposite sides of the vertex. E 
of these portions is caJlc ne. 




tang 



216 



ANALYTICAL GEOMETRY. 




291. To find the equation to the surface of a sphere. The 
sphere is supposed to be generated by the revolution of a semi- 
circle about its diameter. 

If the centre of the sphere be at the origin of co-ordinates, 
then the co-ordinates of any point of the 
sphere, as P, are PM, MN, and AN, and 
we have 

DN 2 =PN 2 =NM 2 + MP 2 ; 
also, 

AD 2 =AN 2 +ND 2 =NM 2 +MP 2 + AN\ 
Hence, putting r for AD, the radius of 
the sphere, we have 

x ^ + f + ^ = r % 
which is the equation of the surface of a sphere. 

292. To find the equation to the surface of a prolate sphe- 
roid. Spheroids are either prolate or oblate. A prolate sphe- 
roid is supposed to be generated by the revolution of an ellipse 
about its transverse axis. An oblate spheroid is supposed to 
be generated by the revolution of an ellipse about its conjugate 
axis. 

Let BCE be an ellipse, and let it be revolved about its trans- 
verse axis ; then any point of the circum- 
ference, as D, in its revolution will de- 
scribe the circumference of a circle. 

Let the origin be placed at the centre of 
the spheroid. The equation of an ellipse 
(Art. 121) is ay + &V = a?b\ 

y = 




or 



a' 



where x represents AN, which is now to 
be represented by z, and y represents NT), 
the radius of the circle described by the 
point D in its revolution. 

Hence ND 2 ^"'" . 

a 



OF SURFACES OF REVOLUTION. 



217 



But* 

hence 



ND 2 = NP 2 = NM 2 + MP 2 =z 2 + ?/ 2 ; 



a 



or a 2 (£ 2 +y 2 ) + 6V = ^ 2 , 

which is the equation of the surface of a prolate spheroid, 

where a is supposed to be greater than b. 

293. To find the equation to the surface of an oblate sphe- 
roid. 

Let the ellipse CBE be revolved about its conjugate axis 
CE ; the point D in its revolution 
will describe the circumference of a 
circle. The equation of an ellipse is 

x - y , 

where y represents AN, which is 
now to be represented by 2, and x 
represents ND, the radius of the 
circle described by the point D in its revolution 

Hence ND 

But 

hence 




fl'}'-flV 



M) 2 =NP 2 =NM 2 +MP 2 =£ 2 +2/ 2 ; 



or b\a? + tf) + a*z* = a*b\ 

which is the equation of the surface of an oblate spheroid. 
The equation of the prolate spheroid is sometimes written 

x* y 7 z 2 

F 2 + F 2+ a 2 = 1 ' 
and that of the oblate spheroid, 



x y z' 
a 2 + a 2 + b 2 ~~ 



:1. 



In both cases a is supposed greater than b. 

If in the equation of either spheroid we make a=b, we shall 



have 



a.' + y' + 3' =7 .*, 



which is the equation of the surface of a sphere (Art. 291). 

Km 



218 



ANALYTICAL GE0METEY. 



294. To find the equation to the surface of a paraboloid. 
A paraboloid is supposed to be generated by the revolution of 
a parabola about its axis. 

Let B AC be a parabola, and let it be revolved about its axis 
AB ; then any point on the curve, as D, 
in its revolution will describe the circum- 
ference of a circle. Let the origin be 
placed at the vertex of the parabola, and 
let the axis of the parabola be the axis of z. 
The equation of a parabola (Art. 85) is 
y^—^ax^ 

y/ where x represents AN, which is now to 

be represented by z, and y represents ND. 

Hence ND 2 :=te. 

But ND 2 = NP 2 = NM 2 + MP 2 = x 2 + y 1 ; 

hence we have x' 2 +y' i =4:az y 

which is the equation of the surface of a paraboloid. 




295. To find the equation to the surface of an hyperboloid. 
An hyperboloid is supposed to be generated by the revolution 
of an hyperbola about one of its axes. 

1st. We will suppose the hyperbola to revolve about its trans- 
verse axis. Let CBD be an hyperbo- 
la, and let it be revolved about its 
transverse axis BE; then any point 
on the curve, as D, in its revolution 
will describe the circumference of a 
circle. Let the origin be placed at 
the centre of the hyperbola, and let 
the transverse axis of the hyperbola 
be the axis of z. 
The equation of an hyperbola (Art. 170) is 

where x represents AN, which is now to be represented by s, 
and y represents ND. 




OF SURFACES OF REVOLUTION. 



219 



Hence 

But 

hence 



ND , =NP , =NM J +MF=aj , +y'; 

to 



or 



which is the equation of the surface generated by revolving an 
hyperbola about its transverse axis. If we suppose both branch- 
es of the hyperbola to revolve, there will be generated two sur- 
faces entirely symmetrical with respect to each other. This is 
therefore called the hyperboloid of revolution of two sheets, 
since it forms two surfaces entirely separate from each other. 

If the asymptotes of the hyperbola also revolve around the 
transverse axis, they will describe the surface of a cone with 
two sheets. The surface of this cone will approach the surface 
of the hyperboloid, and will become tangent to it at an infinite 
distance from the centre. 

2d. We will suppose the hyperbola to revolve about its con- 
jugate axis. Let CBD be an 
hyperbola, and let it be revolved 
about its conjugate axis AE; 
then any point on the curve, as 
D, in its revolution will describe 
the circumference of a circle. 
Let the origin be placed at the 
centre of the hyperbola, and let 
the conjugate axis of the hyperbola be the axis of z. 

The equation of the hyperbola is 




ay + cfO* 



o~ 

where y represents AN, which is now to be represented by s, 
and x represents KD. 

Hence OT>*=^±^-. 

But ND s =NP , =NM , +MP'=jB , +y'; 

hence x'+y'- ~ , 



220 ANALYTICAL GEOMETKY. 

or a^-b\x 2 +f)=-a i b% 

which is the equation of the surface generated by revolving an 
hyperbola about its conjugate axis. As both branches of the 
hyperbola are symmetrical with respect to the conjugate axis, 
each branch in its revolution will describe the same surface. 
This is therefore called the hyperboloid of revolution of one 
sheet, since it forms one uninterrupted surface. 

The equations of the two hyperboloids of revolution are 

2 2 2 

z x y 
sometimes written -^— tz— t5=1, 

a b b ' 

2 2 2 

z x y . 
and __+_+_ =1} 

where the minus sign in each case corresponds to an axis that 
does not meet the surface. 

296. To determine the curve which results from the inter- 
section of a sphere with a plane. Let d represent the distance 
of the intersecting plane from the centre of the sphere ; let the 
origin be at the centre of the sphere, and let one of the co-or- 
dinate planes, as the plane of xy, be parallel to the cutting 
plane ; then every point in the intersecting plane will be given 
by the equation z—d, and we must have 

x*+y*+d*=r% i 
or x*+y 2 =:r*—d% 

which represents all the points on the surface of the sphere 
which are also common to the plane. This equation represents 
a circle whose radius is Vr 2 —d*+ Hence every section of a 
sphere made by a plane is a circle. 

Ex. A sphere w T hose radius is 10 inches is cut by a plane 
whose distance from the centre of the sphere is 6 inches. De- 
termine the radius of the section. 

297. To determine the curve which results from the inter- 
section of a right cylinder with a plane. Every section of a 
right cylinder made by a plane parallel to the base is a circle ; 
we will therefore suppose the section to be made by a plane 



OF SURFACES OF REVOLUTION. 



221 



inclined to the base. Let APB be such a section, and let ABC 
be a section of the cylinder through its axis, and perpendicular 
to the plane of the former section. Draw a 
plane perpendicular to the axis of the cylin- 
der, intersecting the cylinder in a circle whose 
diameter is DE, and intersecting the first 
plane in PM, which will therefore be perpen- 
dicular both to AB and DE, and will be an 
ordinate common to the section and the circle. 
. Let AM=a?, PMrry, AB = 2#, AC=2^; then BM=2#-#. 
We have y 1 = DM . ME (Geom., Bk. IV., Prob. 23, Cor.) ; but 
by similar triangles we have 




AB:AC: 



TX 



: AM : MD, whence MD =- 

a 7 



also 
Whence 



AB : AC : : BM : ME, whence ME = -(2a- a?). 
y^-Qax-x*), 

to 



which is the equation of an ellipse (Art. 129). 

Hence every section of a right cylinder made by a plane in- 
clined to its base is an ellipse. 

Ex. A right cylinder whose diameter is 10 inches, is cut by 
a plane making an angle of 30° with the axis of the cylinder. 
Determine the equation of the elliptic section. 

298. To determine the curve 
which results from the intersec- 
tion of a right cone with a plane. 
Let VBGC be a right cone,V the 
vertex, YH the axis, andBGC the 
circular base. Let AP be the line 
in which the cutting plane meets 
the surface of the cone, and let 
YBHC be a plane passing through 
the axis VII, and perpendicular 
to the cutting plane AMP. AM, 
the intersection of these planes, 




222 ANALYTICAL GEOMETRY. 

is a straight line ; and, since the curve is symmetrical with re- 
gard to it, it is called the axis of the conic section. 

Let DPE be a section parallel to the base ; it will be a circle, 
and DME, its intersection with the plane YBIIC, will be a 
diameter. 

Since the plane DPE and the plane PAM are both perpen- 
dicular to the plane YBHC, MP, the intersection of the two 
former, is perpendicular to the third plane, and therefore to 
every straight line in that plane. It is therefore perpendicu- 
lar to DE and to AM. Draw AF parallel to DE, and ML 
parallel to VB, and let it meet VC in 1ST. 

Let AM=tf, PM=?/, VA=a; let the angle CYH=j3, and 
the angle YAM, which is the inclination of the cutting plane to 
the side of the cone, = 0; then the angle AMN=18O°-0-2/3. 

x sin 

Now AM : ME : : sin. AEM : sin. MAE. whence ME = ~ : 

? cos.p ' 

also, DM=FL=AF-AL=2& sin. /3-AL, and 

AM : AL : : sin. ALM : sin. AML, whence AL ^ Sm '^ ; 

5 cos. p 

therefore DM — 2a sin. ft — x — — — ^— . 

' cos. p 

But by Geom.,Bk. IY., Prob. 23, 

MP 2 = DM. ME; 

x sin. i . x sin. (0 + 2/3) ) - 

which is the equation of the curve resulting from the intersec- 
tion of the cone by a plane. 

Comparing this equation with the equation y*=mx + ?ix' i 
(Art. 234), which represents an ellipse, hyperbola, or parabola, 
according as n is negative, positive, or zero, we see that the sec- 
tion is an ellipse, hyperbola, or parabola according as the co- 
efficient of the last term of the equation is negative, positive, 
or zero. In order to investigate these cases, we will suppose 
the cutting plane to turn about A, so as to make all possible 
angles with the side of the cone. 



OF SURFACES OF REVOLUTION. 223 

299. Discussion of the equation to a conic section. Equa- 
tion (1) of Art. 298 will represent in succession every line 
which it is possible to cut from a given right cone by a plane, 
if we suppose j3 to remain unchanged, while all values are as- 
signed to from to 180°, and all values to a from to in- 
finity. 

Case first. Let = 0; then equation (1) reduces to y 2 = 0. 
This is the equation to the straight line which is the axis of x, 
and we see from the figure that when 0=0 the cutting plane 
becomes tangent to the cone, and the line AM coincides with 
AV. In this case the section is said to be a straight line. 
The same case occurs when 0=:18O . 

Case second. Let 0+2j3<18O°; then sin. (0 + 2/3) will be pos- 
itive ; moreover, sin. is positive so long as is supposed to be 
comprised between and 180°, and cos. 2 j3 is necessarily posi- 

., . sin. sin. (0 + 2/3) . f . . '. m 

tive ; hence — ^ is negative, and equation (1) 

assumes the form y*=mx—nx*, 

which is the equation of an ellipse. We see from the figure 
that in this case the angles YAM and AYF, or ANM, are to- 
gether less than 180°; hence the lines VF and AM, if pro- 
duced indefinitely towards the base of the cone, will meet ; 
that is,, the sectional plane cuts both sides of the cone. Hence 
the section is an ellipse when the cutting plane meets both 
sides of the cone. See fig. Art. 301. 

Case third. In the preceding case the angle may be equal 
to the angle VAF, or 90°-j3,in which case + 2/3 = 9O° + /3, 
and equation (1) reduces to y 2 = 2ax sin. ]3— # 2 , which is the 
equation of a circle (Art. 63). We see that in this case the 
cutting plane is parallel to the base, and hence the ellipse be- 
comes a circle when the cutting plane is parallel to the base 
of the cone. 

Case fourth. Let + 2/3 = 180°; in this case, sin. (0 + 2j3) = O, 
and equation (1) becomes 

# = 2ax sin. tang. /3, 
which is the equation of a parabola (Art. 85). AVe see that in 



224 ANALYTICAL GEOMETRY. 

this case 180° — 0—2/3 — 0; that is, the angle AMN is zero, or 
the cutting plane is parallel to the side of the cone. Hence the 
section becomes a parabola when the cutting plane and the side 
of the cone make equal angles with the base (see fig., Art. 301). 

Case fifth. Let + 2/3>18O°; then sin..(0+2j3) will be neg- 
ative, and — sin. (0 + 2/3) will be positive, and equation (1) as- 
sumes the form y 1 = mx + nx*, 

which is the equation of an hyperbola. We see from the fig- 
ure that in this case the angles VAM and ANM are together 
greater than 180° ; hence the lines YB and AM, though pro- 
duced indefinitely towards the base of the cone, will not meet, 
but if these lines are produced in the opposite direction they 
will meet ; that is, the cutting plane intersects both cones, and 
the curve consists of two branches, one on the surface of each 
cone. 

When 0=180°, the line AM produced returns to the same 
position which it had when 0=0 ; and when becomes greater 
than 180°, the line AM assumes the same positions already de- 
scribed. We therefore obtain all the possible positions of the 
line AM by supposing to be comprised between the limits 
a^nd 180°. 

300. Result of a change in the value of a. The preceding 
results remain unchanged so long as a remains finite. When 
a becomes zero, the cutting plane passes through V, the vertex 
of the cone, and equation (1) becomes 

sin. ^.(1,4-2)3) 

J cos. 2 ]3 v ' 

This equation furnishes three cases : 

Case first Let 0+2/3<18O°; then -sin. (0+2/3) will be 
negative. In this case equation (2) can only be satisfied when 
x—0^ y—Q, which are the equations of a point. A point is 
then to be regarded as a particular case of the ellipse. This 
case happens when the cutting plane, passing through the ver- 
tex V, occupies a position within the angl#BVC'. 

Case second. Let + 2/3 = 180°; then sin. (0 + 2/3) = O, and 



OF SURFACES OF REVOLUTION. 



225 



equation (2) reduces to 2/ 2 = 0. The section then becomes a 
straight line, or it may be regarded as a double line, since the 
equation may be written y=±0. A straight line (or a double 
line) is then a particular case of the parabola. 

Case third. Let + 2/3>18O°; then— sin. (0 + 2/3) will be 
positive, and equation (2) assumes the form 

which represents two intersecting straight lines. This case 
happens when the straight line AM, passing through the ver- 
tex V, meets BC between the points B and C. The cutting 
plane then meets the surface of the cone in two straight lines 
which pass through V. Two intersecting straight lines are 
then to be regarded as a particular case of the hyperbola. 

301. Results of the preceding discussion. It appears from 
the preceding investigation that if a right cone be cut by a 
plane, the section will be 

(1) A parabola when the plane makes 
an angle with the axis equal to half the 
vertical angle of the cone. 

The particular case is a double line. 

(2) An ellipse when the plane cuts 
only one sheet of the cone. 

The particular cases are a point and a 
circle. 

(3) An hyperbola when the plane cuts 
both sheets of the cone. 

The particular case is two straight 
lines which intersect one another. 




302. To determine the curve which results from the inter- 
section of any surface of revolution by a plane. The sections 
of a surface made by the co-ordinate planes are called the prin- 
cipal sections of the surface, and the boundaries of the princi- 
pal sections are called the traces of the surface on the co-ordi- 

K2 




226 ANALYTICAL GEOMETKY. 

nate planes. The equation to a trace is determined by putting 
the ordinate perpendicular to the plane of the trace =0 in the 
general equation (Art. 278). If, then, the cutting plane coin- 
cided with one of the co-ordinate planes, we could easily find 
the trace of the given surface upon that plane, and this would 
be the required curve of intersection. We may make the cut- 
ting plane coincide with one of the co-ordinate planes by a 
transformation of the co-ordinates. In the case of a surface 
of revolution, we may proceed as follows : 

Through AZ, the axis of revolution, draw a plane perpen- 
dicular to the proposed section, and 
call this the plane xz, the origin be- 
ing at A in the plane XAZ. Let 
AX! represent the intersection of the 
cutting plane with the plane xz. The 
lines AX 7 and AY will then be per- 
ri pendicular to each other, and both 

will be in the cutting plane. 

Let P be any point of the curve of intersection, and from P 
draw PM perpendicular to the plane xy, and from M draw MN 
perpendicular to AX. The co-ordinates of P referred to the 
primitive axes are 

z=AN, y=MN, 2=PM. 
Let the point P be now referred to the two axes AX', AY, 
which are in the plane of the given section. Through P draw 
PR perpendicular to AY, and join MR The angle PRM, 
which we will denote by 0, is the angle which the cutting 
plane makes with the plane xy. The co-ordinates of P referred 
to the new axes are 

#' = PR, 2/' = AR=zMN, s'=0. 
In the right-angled triangle PMR we have 

EM = AN = PR cos. PRM, or x = x' cos. 0, 
PM= PR sin. PRM, or z=x' sin. ; 

also we have MN=AR, ovy — y\ 

If the origin be changed to a point in the plane xz whose co- 
ordinates are x — a n y — y z = c n 



OF SURFACES OF REVOLUTION. 227 

these equations become x=a, + x' cos. 0, 

z = c,+ x' sin. Q. 
If tliese values be substituted for x, y, and z in the equation 
of the given surface, the result can only belong to points com- 
mon to the surface and the cutting plane, and will therefore 
represent the required curve of intersection. 

303. To determine the curve of intersection of a plane and 
a prolate spheroid. 

The equation of the given surface (Art. 292) is 

J l a* 

Substituting the values of x, y, and z found in Art. 302, this 
equation becomes 

(a,+x cos. Oy+y 2 +— (c, + x sin. 6y = b% 

or # 2 (cos. 2 0+— sin. 2 0) + y 2 + 2x(-r sin. 6 +a / cos. 0) 
a a 

=v-a;- b X (1) 

' a 
Suppose now the origin to be placed on the surface of the 
spheroid, and in the plane xz. The section of the spheroid by 
the plane xz is equal to the generating ellipse ; hence the co- 
ordinates of the origin must satisfy the equation of the ellipse ; 
that is, we must have 

or y_ a » ^- = o. 

' a 

The second member of equation (1) reduces therefore to 
zero, and the equation is of the form 

y^ — mx— nx 1 , 
and therefore represents an ellipse. If = 0, the equation be- 
comes y* = 2ax—x' i , 
which is the equation of a circle. 

Hence every section of a prolate spheroid by a plane is an 



228 ANALYTICAL GEOMETRY. 

ellipse, except when the cutting plane is perpendicular to the 
axis of revolution, when the section becomes a circle. 

The same is true of the sections of an oblate spheroid. 

Ex. The two axes of a prolate spheroid are 8 and 6, and the 
spheroid is cut by a plane passing through the extremities of 
the axes, and perpendicular to their plane. Required the axes 
of the curve of intersection. Ans. 5 and 3V2. 

304. To determine the curve of intersection of a plane and 
a paraboloid of revolution. The equation of the given surface 
(Art. 294) is x 2 + y' = ^az. 

Substituting the values of x, y, and z given in Art. 302, this 
equation becomes 

(a,+x cos. 6y+y 2 =4ca(c / +x sin. 0), 
or x 2 cos. 2 0-f-?/ 2 + (2# / cos. 0— 4# sin. 0)x=4:ac,—a*. (1) 

Suppose now the origin to be placed on the surface of the 
paraboloid, and in the plane xz; the co-ordinates of the origin 
must satisfy the equation of the generating parabola, and we 
must have a/ = ±ac n or 4ac, — 0/ = 0. 

Equation (1) therefore reduces to the form 

y 2 =zmx— nx*, 
and generally represents an ellipse. If 0=0, the equation be- 
comes y 2 =:2ax—x' i , 
which is the equation of a circle. 

If 0=90°, the equation becomes 

y*=4:ax y 
which is the equation of a parabola. Hence the section of the 
paraboloid by a plane is a parabola, when the plane is parallel 
to the axis of revolution ; it is a circle when the plane is per- 
pendicular to this axis ; and in all other positions of the cutting 
plane the section is an ellipse. 

Ex. A paraboloid whose axis of revolution is 45-?-, and its base, 
or greatest double ordinate, 32, is cut by a plane passing through 
the edge of the base, and meeting the opposite side of the solid 
at the height of 20 above the base. Required the axes of the 
section. Ans. 34.4 and 28. 






OF SURFACES OF REVOLUTION. 229 

305. To determine the curve of intersection of a plane and 

an hyperboloid of revolution. We will suppose the solid to be 

the hyperholoid of two sheets (Art. 295). The equation of the 

b 2 
given surface is x 2 + y 2 — — % z 2 = — b 2 . 

Substituting the values of x } y y and z given in Art. 302, this 

equation becomes 

b 2 

(a y +x cos. 6) 2 +y 2 — -*(c,+x sin. 0) a = — J 9 , 

a 

7 2 72 

or a; 5 (cos. 2 0-- 2 sin. 5 0) +tf- 2z(-r sin. 0-«, cos. 0) 

b'c* 
=-#-*-«/• (1) 

If we place the origin on the surface of the hyperboloid, and 
in the plane a&, the second member of this equation reduces to 
zero, and the equation is of the form 

?/ 2 =m#— nx 2 . 
If 0=0, the equation becomes 

y 2 =2ax— x 2 , 
which is the equation of a circle. 
If 0=90°, the equation becomes 

y 2 ~(x 2 +2c,x), 

which is the equation of an hyperbola. 

If tang. 0=t, the equation reduces to 

y 2 = 2x(c / cos. cotang. 0— #, cos. 0), 

which is the equation of a parabola. 

a 2 a 2 

If tang. 2 0<Ta-, the curve is an ellipse; if tang. 2 0>^, the 

curve is an hyperbola. 

In every case the section of the hyperboloid by a plane is 
similar to the corresponding section of the cone formed by the 
revolution of the asymptotes of the hyperbola (Art. 295). 



230 ANALYTICAL GEOMETRY. 

306. Summary of the preceding results. The equation to 
the surface of an oblate spheroid (Art. 293) may be written 

x* ?/ £ 

-i+^+r* =1 5 (1) 

a a b ' w 

and that of a prolate spheroid, 

The equation to the surface of an hyperboloid of one sheet 

r 2 ?y 2 s 2 
(Art. 295) is ^+|i-ji=l, (3) 

and that of an hyperboloid of two sheets is 

The equation to the surface of a right cone (Art. 290) is 
x*+y*—z* tang. 2 ^ = ; 

if we divide oy a Q , and put V for - — ^-5-, the equation becomes 

-5+^-5=0. (5) 

a a v J 

The equation to the surface of a paraboloid (Art. 294) is 

#* + 2/ 2 — 4a£=0; 

if we divide by a*, and put 5 for 7, the equation becomes 

^!+^_5; = o. ( 6) 

a a b w 

In each of these six equations the coefficients of a? and y* 
are equal, which shows that for each of these solids a section 
perpendicular to the axis of z is a circle. 

307. More general form of the preceding equations. If we 

suppose the coefficients of x* and y* in either of these equations 

to be unequal, we shall have a new equation similar in form to 

the preceding, but representing a more complex surface. The 

x* ?/ 2 2 
equation _ + - + - = l (1) 

represents a surface similar in some respects to that of the 



OF SURFACES OF REVOLUTION. 231 

spheroid, but its intersection with a plane perpendicular to the 
axis of z is an ellipse instead of a circle. All sections made 
by parallel planes are similar ellipses, and the surface is closed 
in every direction. This solid is called an ellipsoid, and has 
three unequal axes. When two of the axes are equal to each 
other it is called an ellipsoid of revolution, because it may be 
generated by the revolution of an ellipse about one of its axes. 

2 2 2 

The equation a^ + l/~"? = 1 ^ 

represents a surface like the hyperboloid of one sheet, except 
that the sections perpendicular to the axis of z are ellipses in- 
stead of circles. 

2 2 2 

£P ?/ Z 

So also the equation -5 + ~ — - 2 = — 1 (3) 

CL C 

represents a surface like the hyperboloid of two sheets, but the 
sections perpendicular to the axis of z are ellipses. 

2 2 2 

__ . x y z ^ 

The equation -, + p — -5 = (4) 

represents a conical surface, but the cone has an elliptic base 
instead of a circular one. 

The equation — + ^—- = (5) 

represents a surface like the paraboloid of revolution, except 
that a section perpendicular to the axis of z is an ellipse in- 
stead of a circle. This solid is called an elliptic paraboloid. 

Each of these surfaces may be conceived to be derived from 
the corresponding surface of revolution by increasing or dimin- 
ishing the values of y in a constant ratio, in the same manner as 
oblate and prolate spheroids may be derived from the sphere 
by multiplying the values of y by a constant factor, or as the 
ellipse may be derived from the circle by multiplying the values 
of y by a constant factor. 

^308. Surface of a cone asymptotic. The conical surface 

represented by the equation 

222 

X V Z r, 



—jJL __ i 



232 ANALYTICAL GEOMETRY. 

is asymptotic on the one side to the hyperboloid of one sheet 

2 2 2 

Cu 7J 2/ 

whose equation is -5+za — -*=1, 

and on the other side to the hyperboloid of two sheets whose 
equation is 

There is also a similar relation between the equations of two 
conjugate hyperbolas and the equation of their asymptotes. 
The equation of an hyperbola (Art. 170) may be written 

and the equation of its conjugate hyperbola (Art. 179) is 

a' V~ ' 
while the equation of their asymptotes (Art. 214) is 



or 






GENERAL EQUATION OF THE SECOND DEGREE, ETC. 233 



SECTION V. 

GENERAL EQUATION OF THE SECOND DEGREE BETWEEN THREE 

VARIABLES. 

309. The general equation of the second degree between 
three variables is of the form 

ax*+bxy+cy 2 + dz* + exz-]-fyz+gx-\-hy+Jcz + l=0. (1) 
We may transform this equation into another, in which the 
axis of z remains unchanged, by employing the equations of 
transformation for plane co-ordinates (Art. 55), and we shall 
have z — z' 

x—x f cos. 0— y' sin. 9 
y~x' sin. Q + y f cos. 0. 
If we substitute these values of the variables in equation (1), 
the only terms in the resulting equation which can contain the 
product x'y' will come from the three terms ax* + bxy + cy\ 
The term containing xy may therefore always be made to dis- 
appear from equation (1) by the method explained in Art. 230. 
So, also, the term containing xz may always be made to dis- 
appear by a new transformation, in which the new axis of y 
remains unchanged ; and in the same manner the term con- 
taining yz may be made to disappear. Hence equation (1) can 
always be transformed into an equation of the form 

Ax' + By' + Cz'+Bx+Ey+Fz+G^O. (2) 

If, in equation (2), neither A, B, nor C is zero, we may, as in 

Art. 229, cause the terms containing the first powers of x, y, 

and z to disappear by changing the origin of the co-ordinates, 

and the equation will be reduced to the form 

Lz 2 +H*/ 2 +IsV+P=0. (3) 

310. Classification of the surfaces represented by the equa- 
tion (3). In discussing equation (3) we must suppose each of 
the coefficients to be either plus or minus, and we must also 
consider the case in which P reduces to zero. Now two of the 



234 ANALYTICAL GEOMETRY. 

coefficients L, M, and 1ST must always have the same sign ; we 
will suppose that L and M have the same sign, and will make 
these signs positive. We may then have the six following 
cases : 

1. When N is plus and P minus. Equation (3) will then 
take the form Lz 2 + My 2 + Ns 2 - P = 0. 

P P P 

If we divide by P, and put a* = y- 3 ^—^ an d c* = t^j we shall 

x' f z 2 , 
have "~a+w+-2=l 5 

which is the equation of the surface of an ellipsoid (Art. 307, 
eq.l). 

2. When N is phis and P plus. Equation (3) will then be- 
come L# 2 + M?/ 2 +]NV + P=0, 

in which all the terms are positive. Hence the equation can 
not be satisfied for real values of the variables, and therefore 
the surface becomes imaginary. 

3. When N" is plus and P is zero. Equation (3) will then 
become Lx' + My* + Ns 2 = 0, 

which can only be satisfied by the values 

oj = 0, y=®, z = 0; 
and hence this supposition reduces the surface to B,jpoi?it,\iz. y 
the origin. 

4. When N is minus and P is minus. Equation (3) will then 

become L^ 2 + My 2 - IsV - P = 0. 

P P P 

If we divide by P, and put a a =y 3 ^^lyf? an ^ c<i= w we s ^ ia ^ 

# 2 y 1 2 2 
have —^+ Ti— -1 = 1, 

which represents the surface of an hyjperboloid of one sheet 
(Art. 307, eq. 2). 

5. When N is minus and P is plus. Equation (3) becomes 

Lz 2 + Mt/ 2 -1sV + P = 0. 
Substituting as before, we have 









GENERAL EQUATION OF THE SECOND DEGREE, ETC. 235 

which represents the surface of an liyperboloid of two sheets 
(Art. 307, eq. 3). 

6. When N is minus and P is zero. Equation (3) becomes 
LB'+My'-Ne'zzOj 

which by substitution becomes 

x y z 

which represents the surface of a cone having an elliptic base 
(Art. 307, eq. 4). 

311. Particular cases of the general equation. If both 
terms containing one variable, as z, are wanting from eq. (2), 
Art. 309, that is, if C and F are zero, all sections of the surface 
perpendicular to the axis of z are equal to each other, since the 
equation is independent of z. The common equation of these 
sections is Ax* + By 2 + Dx + Ey + Gr = 0, 

which may represent either of the conic sections (Art. 233). 
This surface is called a cylindrical surface, and may be de- 
scribed either — 

1. By the above-named conic section moving always parallel 
to itself and along a right line parallel to the axis of z, or 

2. By a straight line which moves along the conic section, 
and in all of its positions is parallel to the axis of z. 

The conic section is called the base of the cylinder, and the 
cylinder is called circular, elliptic, hyperbolic, or parabolic, ac- 
cording to the nature of the base. 

When the equation Ax* + By* -\-Dx + T£y+G = represents 
two straight lines (Art. 233), the cylindrical surface becomes 
two planes, which may intersect or be parallel, or may coincide 
as a double plane. 

When two of the three coefficients A, B, and C in eq. (2), 
Art. 309, are zero, as B and C, one of the corresponding terms 
~Ey and Yz may be made to disappear by a transformation 
in which x remains unchanged, but the axes of y and z are 
changed in the plane yz, and the resulting equation is that of 
a cylinder, as above. 



236 ANALYTICAL GEOMETRY. 

312. Elliptic and hyperbolic paraboloids. The only re- 
maining case of eq. (2), Art. 309, is when two of the coeffi- 
cients, as A and B, are finite, and the third is zero. The first 
powers of x and y can then be made to disappear by changing 
the origin of x and ?/, and the constant term may be made to 
disappear by changing the origin of z. The equation will then 
become A^+B^+F^O, 

x* li 1 % 
which may be written -5 + ts + - = 0. 

If A and B have like signs, the surface is that of an elliptic 
paraboloid ; if A and B have unlike signs, every cross section 
perpendicular to the axis of 2 becomes an hyperbola, and the 
surface is called an hyperbolic paraboloid. 

313. How an elliptic or hyperbolic paraboloid may be de- 
scribed. A parabola may be regarded as the limiting case of 
an ellipse, one vertex of which is fixed, and the other is re- 
moved to an indefinitely great distance. So, also, the elliptic 
paraboloid may be regarded as an ellipsoid, one of whose axes 
has been indefinitely incro&sed, while one vertex of that axis 
remains fixed. 

The elliptic paraboloid may be regarded as described by one 
parabola moving upon another. Thus, let the plane of one 
parabola be at right angles to the plane of another; let the 
axes of the two parabolas coincide, and the concavities be 
turned in the same direction. Then, if one of the parabolas 
move so as to be always parallel to itself and to have its vertex 
upon the fixed parabola, the surface described by the movable 
parabola will be an elliptic paraboloid. 

But if the concavities of the two parabolas are turned in op- 
posite directions, the corresponding surface thus described will 
be an hyperbolic paraboloid. 

314. Section of a surface of the second degree by a plane. 
Every intersection of a plane with a surface of the second de- 
gree is either a straight line or one of the conic sections. 






GENERAL EQUATION OF* THE SECOND DEGREE, ETC. 237 

For by one or two transformations of co-ordinates like those 
of Art. 309 we can refer the surface to a new system of co-or- 
dinates, one of which, as 2, will be parallel to the given inter- 
secting plane. In these transformations it is evident that the 
degree of the equation can not be increased, since the values 
substituted for x, y, and z are always of the first degree. If 
now we substitute for z in the transformed equation the dis- 
tance of the intersecting plane from the plane a??/, we shall 
have an equation between x and y, which is the equation of 
the intersection of the plane and surface. The degree of this 
equation does not exceed the second, and therefore (Art. 233) 
the curve must be either a straight line or a conic section. 

The conic section may, however, in special cases, break up 
into two lines, as shown in Art. 233. 



APPENDIX. 



ON THE GRAPHICAL REPRESENTATION OF NATURAL LAWS. 
TnE mutual dependence existing between any two or more 
variable quantities may be exhibited by means of curve lines. 
If, for example, we have a large collection of meteorological 
observations showing the temperature at any place for each 
hour of the day, the nature of the relations or laws expressed 
by these numbers may be represented by curve lines. Such a 
mode of representation frequently renders these laws perfectly 
obvious, and sometimes suggests relations which might easily 
have been overlooked in a large mass of figures arranged in 
tables. There is a great variety in the modes by which this 
representation may be effected. The following are some of 
the methods most frequently employed : 

I. Relations of two variables expressed by rectangular co- 
ordinates. If on a horizontal line we set off distances propor- 
tional to the values of one of the two variables, regarding these 
as abscissas, and from the several points of division erect per- 
pendiculars whose lengths are proportional to the correspond 
ing values of the other variable, and then draw a continuous 
curve line through the extremities of these perpendiculars, 
this curve line may be regarded as representing the relation 
between the two variables. The cases of this nature most fre- 
quently occurring are those in which time is one of the varia- 
bles, and this is usually laid off upon the axis of abscissas. 

Ex. 1. Diurnal change of temperatwe. Let it be proposed 
to construct the curve which represents the relation between 
the different hours of the day and the corresponding mean 
temperature at a given place. The following table shows the 



240 



APPENDIX. 



mean temperature at New Haven for each hour of the day, as 
deduced from a long series of observations : 



Hour. 


Temp. 


Hour. 


Temp. 


Hour. 


Temp. 


Hour. 


Temp. 


Midnight 


45°.0 


6 A.M. 


43°. 1 


Noon. 


55°. 3 


6 P.M. 


52°.0 


1 A.M. 


44 .3 


7 " 


44 .6 


1P.M. 


56 .1 


7 " 


50 .2 


2 " 


43 .6 


8 " 


46 .9 


2 " 


56 .5 


8 " 


48 .7 


3 " 


43 .1 


9 " 


49 .7 


3 " 


56 .3 


9 " 


47 .5 


4 " 


42 .7 


10 " 


52 .2 


4 " 


55 .4 


10 " 


46 .5 


5 " 


42 .6 


11 " 


54 .0 


5 " 


53 .9 


11 " 


45 .7 





— 
















rS 








































y 


' 


S 


X 


















/ 






\ 
















/ 










\ 














/ 










\ 












j 


/ 












X, 










/ 
















^> 



















































m't2h. 4 6 8 10 noon 2h. 4 



In order to represent these observations by a curve line, we 

draw upon a sheet of pa- 
per a horizontal line, and 
divide it into twenty-four 
equal parts, to represent 
the hours of the day, and 
through these points of 
s lom't division we draw a system 
of vertical lines. Upon each of these vertical lines we set off a 
distance proportional to the height of the thermometer for the 
corresponding hour, and then connect all the points thus de- 
termined by a continuous line. The curve thus formed repre- 
sents the mean motion of the thermometer at New Haven for 
the different hours of the day, and, if constructed with proper 
care, and upon a scale of suitable size, may supply the place of 
the numbers from which it was derived, the temperatures being 
indicated by the numbers on the left of the diagram. In order 
to avoid confusion, the ordinates in the diagram have only been 
drawn for the alternate hours. 

We readily perceive from the figure that on each day there 
is a maximum and a minimum of temperature, the maximum 
occurring generally about two hours after noon, and the mini- 
mum about an hour before the rising of the sun. We see, also, 
that the temperature is increasing during nine hours of the day, 
and decreasing during the remaining fifteen hours of the day. 
This curve readily shows us the two periods of the day when 
any given temperature is attained ; as, for example, a tempera- 
ture of 50°, 52°, etc. It also shows, not only the mean tern- 






AITENDIX. 



241 



perature at the hours of observation, but also for any time 
intermediate between these hours; as, for example, for each 
half hour, or quarter hour, etc. 

Ex. 2. Annual change of temperature. In the same man- 
ner we may construct the curve representing the connection 
between the different months of the year and the correspond- 
ing temperature at a given place. We draw a horizontal line, 
and divide it into twelve equal parts, to represent the months 
of the year, and through these points of division draw a system 
of vertical lines, upon which we set off distances proportional to 
the heights of the thermometer for the corresponding months. 

The following table shows the mean temperature of New 
Haven for each month of the year, as deduced from a long 
series of observations. It also shows the average maximum 
temperature of each month, and the average minimum tem- 
perature of each month : 





Mean 


Maximum 


Minimum 




Mean 


Maximum 


Minimum 




Temp. 


Temp. 


Temp. 




Temp. 


Temp. 


Temp. 


Jan... . 


26°.5 


49°.6 


-1°.0 


July.. 


71°.7 


90°. 8 


52°.8 


Feb.... 


28 .1 


51 .3 


+ 1 .0 


Aug.. 


70 .3 


88 .6 


50 .0 


March. 


36 .1 


61 .6 


10 .7 


Sept.. 


62 .5 


83 .6 


37 .6 


April. . 


46 .8 


72 .6 


25 .4 


Oct... 


51 .1 


73 .2 


26 .7 


May... 


57 .3 


81 .3 


35 .5 


Nov.. 


40 .3 


63 .2 


17 .7 


June . . 


67 .0 


89 .3 


45 .9 


Dec. 


30 .4 


53 .1 


4 .5 



In the annexed fig- 
ure, the middle curve 
line shows the mean 
temperature of each 
month of the year, ac- 
cording to the preced- 
ing observations,while 
the upper curve shows 
the average maximum 
temperature, and the 20 
lower curve the aver- 10 
age minimum temper- 
ature for each month 
of the year. 



90 c 



so 



CO 



40 



so 





































/« 


A*IM!, 


M~^ 


















/ 








\ 
















/ 










\ 












/ 












\ 












/ 






^MEA 


N > 


\ 




\ 








/ 






/ 






\ 




\ 








/ 




/ 








s 


V 


s 


v 




^A 






/ 










\ 




s 








/ 




/ 


'minii 


rfUWv 




\ 












/ 




/ 






\ 












/ 




/ 








\ 












/ 




/ 








s 


V 






s^___ 






A 












\ 




\ 








/ 














y 










/ 














\ 








/ 
















s 


V 






/ 


















\ 




/ 




















\ 























M J J 



O Is D 



242 APPENDIX. 

These curves inform us that at Is ew Haven the warmest 
months of the year are July and August, and the maximum for 
the year occurs near July 24th. The coldest month of the 
year is January, and the minimum for the year occurs near 
Jan. 21st. The difference between the minimum and the max- 
imum for each month is greater in the cold months than in the 
warm months. Various other particulars respecting the con- 
nection between the temperature and the season of the year 
are also exhibited by the figure more palpably than by a col- 
umn of numbers in a table. 

The same mode of representation may be employed to ex- 
hibit the relation between the height of the barometer and the 
hour of the day or the season of the year; also for the amount 
of vapor in the atmosphere, the force of the wind, the fall of 
rain or snow, the prevalence of cloud or fog, the intensity of 
atmospheric electricity, the declination or dip of the magnetic 
needle, or the intensity of terrestrial magnetism, or, indeed, any 
natural phenomenon which depends on the course of the sun. 

Ex. 3. Display of November meteors. On the morning of 
Nov. 14, 1866, a remarkable display of meteors was witnessed 
in England, and the sudden increase, as well as the equally sud- 
den decline in the number of meteors, is exhibited by a curve 
line much more strikingly than could be done by a simple nu- 
merical statement. For this purpose we draw a horizontal 
line, and divide it into equal parts, to represent the hours of 
observation, and through the points of division we draw a sys- 
tem of vertical lines. On these vertical lines we set off dis- 
tances proportional to the number of meteors counted eacli 
minute, and through the points thus determined we draw a 
continuous curve line. The numbers on the left margin of 
the figure on the opposite page indicate the number of meteors 
visible each minute. From the diagram we perceive that be- 
fore midnight the number of meteors did not exceed 5 per 
minute, but soon after midnight the number rapidly increased, 
and at lh. 20m. exceeded 120 per minute ; by 2 A.M. it had de- 
clined to 40 per minute, and by 3 A.M. to 10 per minute. 



AITENDIX. 
taPIU. a MIDNIGHT IM. 2 3 4 5 



243 











A 








120 






( 


\ 
























100 








\ 








80 








\ 




























!\! 
















1 


u 








40 






r 


1 












1 


u 


i 












/ 




V 
















^ 











L- — -H 








^ — . 



A similar mode of representation may be advantageously 
employed to exhibit the results of a large mass of observations, 
even though we have no previous knowledge of the laws which 
govern their changes. We may thus exhibit the influence of 
the day or the season of the year upon mortality ; we may ex- 
hibit the average number of deaths at different ages ; or we 
may exhibit the fluctuations in the price of any article of mer- 
chandise, as wheat, cotton, gold, etc. 

Ex. 4. Annual change in the depth of rivers. The depth of 
the water in the Mississippi River fluctuates greatly with the 
season of the year. During the early part of autumn the water 
is usually lowest, and it is highest some time in the spring or 
the early part of summer. The figure on the following page 
shows the average result of twenty-three years of observations 
on the river at Katchez, Miss. The months are shown at the 
bottom of the figure, while the depth of water is indicated by 
the numbers on the left margin. 

We see from this figure that the water is usually lowest in 
October, when its depth is only 12.5 feet. From this time the 
water rises pretty steadily to the first of May, when the depth 
amounts to 48.3 feet, from which time it declines pretty steadily 



244 



APPENDIX. 



till the following Oc- 
tober. There are, 
however, two small- 
er maxima which are 
well marked,viz.,one 
about the 1st of Feb- 
ruary, and the other 
about the middle of 
June. These great 
fluctuations of the 
Mississippi are due 
not so much to an 
excess of rain near 
the time of maxi- 
mum height as to 
the melting of the 
snow accumulated 
upon the numerous 
tributaries of this 
river. 

Ex. 5. Velocity of 
the current of a riv- . 



Dec. Jan. Feb. Mar. Apr. May Jun. Jul. Aug. Sep. Oct. Xov. CT at different deaths. 



It has been found by experiment that the velocity of the cur- 
rent in rivers varies sensibly with the depth. This may be 
shown by means of floats immersed to different depths in the 
water. The following is one mode of performing the experi- 
ment : A keg 15 inches high and 10 in diameter, without top 
or bottom, is ballasted with lead so as to sink and remain up- 
right in the water ; the keg is attached by a small cord to a 
mass of cork 8 inches square and 3 inches thick, and a small 
flag is supported by the cork, in order that it may be more 
readily observed at a distance. By varying the length of the 
cord, the keg may be made to sink to any required depth, and 
its size is so much greater than that of the surface-float that 
the latter does not sensibly affect the rate of movement. 







APPENDIX. 



245 



Surface Z.r> 



The apparatus being placed in the water, its rate of motion 
is determined by observers stationed on the bank of the river 
at known distances from each other, and watching the progress 
of the float by means of theodolites. 

The curve line on the annexed figure shows the result of 
experiments made on the cur- 
rent of the Mississippi near New 
Orleans. The numbers on the 
left margin show the depth of 
the keg, expressed in tenths of 
the entire depth of the river, the 
mean depth of the water being 
86 feet. The numbers at the 
top of the figure show the ve- 
locity of the current, expressed °- : - 
in miles and tenths of a mile 
per hour. 

TTe see from the figure that 
the velocity at the surface is 
3.74 miles per hour ; the velocity increases as we descend, until 
we reach a depth about one third that of the river, where the 
velocity amounts to 3.S4 miles per hour, while below this depth 
the velocity diminishes, and at the bottom of the river is re- 
duced to 3.47 miles per hour. 

Ex. 6. Average duration of human life. The average du- 
ration of life may be deduced from tables which show the 
number of deaths which occur each year out of a given num- 
ber of individuals. If there were a million of births in the 
year 1770, and we had a record of the number of deaths out 
of this company for each year to the present time, we could 
construct a table showing the average duration of life for each 
age. The average duration of life for a person of a certain 
age is understood to be the average number of years which 
the survivors of that age should live. The duration of life is 
different in different countries. The curve line in the follow- 
ing figure shows the average duration of life as deduced from 




Euttcra 



246 



APPENDIX. 



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10 15 20 25 CO 35 40 45 50 55 CO 65 70 75 SO 85 i)0 95 100 

observations made at Carlisle, Eng. The numbers at the bot- 
tom of the figure show the age of the individual from to 100 
years, and the numbers on the left margin show the average 
duration of life. This average duration of the life of individ- 
uals after any specified age is called the expectation of life. 

We see from the figure that the average duration of life 
for an infant just born is 38 years. If the child survives, its 
expectation of life increases for a few years, and attains its 
maximum at the age of 5, when the average duration of life is 
51 years. After this age the average duration of life dimin- 
ishes steadily and pretty uniformly until death. At the age of 
25 the average duration of life is 38 years, at 50 it is 21 years, 
at 75 it is 7 years, and at 100 it is 2 years. 



II. Relations of several variables depending upon a com- 
mon variable. When we have several variable elements de- 
pending upon a common variable, we may graduate a horizon- 
tal line to represent successive values of the common variable, 
and then construct a number of curve lines to represent the 
changes in each of the other variables. A comparison of the 
different curves will show not only the relation of each variable 
to the common variable, but also the mutual relation of the 
several variables. 



APPENDIX. 



247 



Ex. 1. Temperature beloio the earth's surface. Suppose we 
wish to discover how the diurnal and annual changes of tern- 
perature are modified by depth below the surface of the earth. 
For this purpose we require observations of temperature made 
at different depths below the earth's surface, and continued at 
least throughout an entire year. Such observations have been 
made at several places in Europe. Thermometers w T ith very 
long stems have been buried at depths of 24, 12, 6, and 3 French 
feet, and 1 inch, and the observations have been continued for 
many years. The annexed figure presents a summary of such 

Observations COIltm- Apr. May Jun. Jgl^Ang. Sep. Oct. IVov. Dec. Jan. Teb.Mar.Ap r. 

ned for 14 years at ' 
Greenwich, the 
months being given 
at the top of the fig- 
ure, and the temper- 
atures on the left 
margin. 

We perceive from 
the figure that at a 







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depth of about 6 feet 5 o 
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the annual range of temperature is less than one third, and at 
the depth of 24 feet it is only one ninth what it is at the sur- 
face. We also perceive that the highest temperature of the 
year occurs later and later as we descend below the surface of 
the earth. At the depth of 12 feet the maximum temperature 
of the year occurs about the last of September, and the mini- 
mum about the last of March, while at the depth of 24 feet the 
maximum occurs about the first of December, and the minimum 
about the first of June. 

Ex. 2. Declination of the magnetic needle and the solar 



21S 



APPENDIX. 



spots. The surface of the sun often exhibits black spots of 
irregular form and variable size. The number of these spots 
varies greatly in different years ; sometimes the sun is entirely 
free from them, and continues thus for months together, while 
some years the sun is never seen entirely free from spots. The 
curve in the lower part of the annexed figure presents a sum- 



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mary of observations of the spots for a period of 04 years, the 
dates being given at the bottom of the figure, while the fre- 
quency of the spots is exhibited on the left margin by a scale 
of numbers extending from to 100. We readily perceive 
that the spots are subject to a certain periodicity, the number 
of the spots increasing for 5 or 6 years, and then decreasing 
for several years, showing alternate maxima and minima. The 
maxima occurred in 1817, 1830, 1837, 1818, and 1860, while 
the minima occurred in 1810, 1823, 1833, 1843, 1856, and 1867. 
A magnetic needle, when freely suspended and carefully ob- 
served from hour to hour, exhibits a small daily oscillation va- 
rying from 5 r to 15 7 , The extent of this oscillation varies with 
the season of the year, and the mean annual range varies from 
one year to another. The curve in the upper part of the above 
figure shows the results of observations of the magnetic needle 



APPENDIX. 



249 



made in Europe for a period of G4 years, the dates being shown 
at the bottom of the figure, and the mean daily average of the 
needle being shown by numbers on the left margin, which rep- 
resent minutes of arc. 

We see from the figure that the range of the needle, which 
was only 6' in 1810, had increased to 8' in 1818, had decreased 
again to about 6' in 182-1, and increased to 10' in 1829, etc. In 
other words, the annual range of the magnetic needle shows 
alternate maxima and minima, and the times of these maxima 
correspond remarkably with the maxima of the solar spots, 
suggesting the idea that the two phenomena are dependent 
upon a common cause. Such a mode of representation by 
curve lines is well calculated to show the connection between 
two different classes of phenomena. 



III. Relations of two variables expressed by polar co-ordi- 
nates. The relations between two variable elements may be 
expressed by means of polar co-ordinates, and this method is 
generally to be preferred when one of the variables denotes 
direction ; for example, if one of the variables is the direction 
of the wind, and the other variable is the corresponding mean 
height of the barometer, or thermometer, or hygrometer* For 
example, suppose we wish to show the dependence of the tem- 
perature of the air upon the direction of the wind. 

Ex. 1. Influence of the wind on temperature. From a com- 
parison of several years of observations, it has been found that 
at Xew Haven the temperature of the air during the prevalence 
of winds from the eight principal points of the compass differs 
from the mean temperature of the year by the quantities shown 
in the annexed table : 



Wind. TL-mperature. 


Wind. 


Temperature. 


North 

Northeast 

East 

Southeast 


— 2°. 7 

-o !g 

+ .5 
+ 1 .2 


South 

Southwest 

West 

Northwest 


+3°.2 
+ 4 .0 
-1 .1 
-4 .5 



In order to represent these results by a curve line, we draw 

L2 



250 



APPENDIX. 




eight radii inclined to each oth- 
er in angles of 45°, to represent 
the directions of the wind. "With 
the point A as a centre, we draw 
a series of equidistant circum- 
J E f erences, to represent differences 
of temperature, and then, having 
selected one of these to represent 
the mean temperature of New 
Haven, we set off upon the eight 
radii distances proportional to 
the numbers in the preceding table. When the numbers are 
negative, we set them off towards the centre of the circle ; 
when they are positive, we set them o&from the centre. The 
curve line passing through the eight points thus determined 
shows the influence of the wind's direction upon the tempera- 
ture of the air. We perceive that the highest temperature ac- 
companies a wind from S. 33° W., and the lowest temperature 
corresponds to a wind from the point N. 40° W., the mean dif- 
ference in the temperature of these two winds being 8°.7. 

Ex. 2. Direction of the prevalent wind. The prevalent wind 
at any station may be graphically represented by means of polar 

co-ordinates. Suppose we have a 
long series of observations of the 
wind from which we deduce the 
number of times the wind was ob- 
served to blow from the north 
point; also the number of times 
it blew from the northwest, the 
number of times from the west, 
and so on, for 8 or 16 points of the 
compass. We draw two lines at 
right angles to each other to rep- 
resent the cardinal points, and also 
other lines to represent the interme- 
diate directions. From the point 



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s.w.' ^1 




Ns.E. 



E. 



APPENDIX. 251 

of intersection we set off on these lines distances correspond- 
ing to the relative frequency of the winds from these different 
points of the compass. The curve line passing through the 
points thus determined may be regarded as showing the prev- 
alent wind for that station. 

The preceding figure shows the results of observations made 
during the month of January for several years at Wallingf ord, 
near New Haven. We see that the prevalent wind is almost 
exactly from the north, but that winds from the S.S.W. are also 
of frequent occurrence. 

This mode of representation is valuable when we wish to 
exhibit the peculiarities of a large number of stations. The 
eye is thus able at a glance to detect characteristic peculiarities 
which might be easily overlooked in a large collection of nu- 
merical results. 

Ex. 3. Diurnal change in the direction of the wind. An- 
other mode of representation, bearing some resemblance to the 
preceding, may be advantageously employed to denote the con- 
nection between the hour of the day and the corresponding 
direction of the wind. Suppose, from a long series of observa- 
tions, we have determined the mean direction of the wind for 
each hour of the day. Having drawn two lines at right angles 
to each other to represent the cardinal points of the compass, 
we begin with the observation for the first hour, and draw a 
line of any convenient length to represent the wind's direction 
at that hour ; from the extremity of this line we draw a line 
of the same length as before, to represent the wind's direction 
at the second hour, and in the same manner we set off the di- 
rections of the wind for each of the twenty-four hours. "We 
thus construct a broken line, which may be regarded as repre- 
senting the average progress of a particle of air for each hour 
of the day, supposing the wind's velocity to have been the same 
at all hours ; or, if we have observations showing the wind's 
velocity for each hour, we may make the portions of the curve 
which represent the wind's direction for the different hours rep- 
resent not only its direction, but, at the same time, its velocity. 



252 



APPENDIX. 



The annexed figure shows the mean di- 
rection of the wind at New Haven for the 
different hours of the day during the month 
of August. We perceive that early in the 
morning the average direction of the wind 
for this month is from the north, while dur- 
ing the afternoon its average direction is 
from the south, and about 10 A.M. the w r ind 
veers from N. to S., going round by the east. 
This diurnal change in the wind's direction 
constitutes what is commonly known by 
the name of a " land and sea breeze." 
The change in the wind's direction for the other months of 
the year may be represented in a similar manner. 




IV. Contour lines and geographical distribution. If it is 
required to represent upon a map the undulations in the surface 
of a tract of land, we suppose the surface of the ground to be 
intersected by a number of horizontal planes at equal distances 
from each other, and we delineate on paper the curve lines in 
which these planes intersect the surface. 

Ex. 1. Survey of an undulating surface. This method will 
be understood from the annexed figure, which represents a 



B 




APrENDIX. 



253 



tract of broken ground divided by a stream, EF. The ground 
is supposed to be intersected by a horizontal plane four feet 
above F, the lowest point of the tract, and this plane intersects 
the surface of the ground in the undulating lines marked 4, one 
on each side of the stream. A second horizontal plane is sup- 
posed to be drawn eight feet above F, and this intersects the 
surface of the ground in the lines marked 8. In like manner, 
other horizontal planes are drawn at distances of 12, 16, etc., 
feet above the point F. The projection of these lines upon 
paper shows at a glance the outline of the tract. 

Ex. 2. Depth of water in a harbor. If we have soundings 
showing the depth of water at numerous points of a harbor, 
the results may be delineated on paper in a similar manner. 
We draw a curve line joining all those points where the depth 
of water is the same — for example, 10 feet. We draw another 
line connecting all those points where the depth of water is 20 
feet ; also other lines for 30 feet, 40 feet, etc. 

The accompanying figure represents a portion of New York 
Harbor, and the dotted lines show depths of 20, 40, and 60 feet 



of water. 



We see that along the channel of the North River 




254 APPENDIX. 

there is every where a depth of at least 40 feet, but in passing 
from the North River to East River there are obstructions 
where the depth of water is only 20 feet. 

A similar principle is now very extensively employed to rep- 
resent almost every variety of variable quantity depending 
upon geographical position. In many cases the representation 
is greatly assisted by variations in the depth of shading, or by 
varieties of color, etc. The following examples will afford 
some idea of this method. 

Ex. 3. Lines of equal mean temperature. We draw upon a 
map of the earth a curve line connecting all those places whose 
mean temperature is the same — for example, 80°. As it may 
happen that we have no station wiiose observed temperature is 
exactly 80°, we select two adjacent stations, at one of which 
the temperature is a little less than 80°, and at the other a 
little greater; we then divide the interval between them in 
the same ratio as the differences between the observed temper- 
atures and 80°. The point thus determined we call a point of 
80° temperature. In the same manner we determine as many 
points of this line as practicable. Next we draw a line con- 
necting all those places whose mean temperature is 70°, G0°, 
50°, etc. The figure on the opposite page exhibits such a sys- 
tem of lines for nearly the entire globe. Maps of this kind, 
when carefully constructed, give a much clearer idea of the 
distribution of heat on the earth's surface than can be done by 
any system of numbers arranged in tables. 

In like manner we may draw lines representing the mean 
temperature of different places for any month of the year, or 
we may draw lines to represent the temperatures observed for 
any given day and hour, thus enabling us to study the actual 
distribution of temperature at any instant of time. 

Ex. 4. Lines of equal atmosp7ieric pressure. We may draw 
upon a map of the earth a curve line connecting all those places 
where the mean pressure of the air, as shown by a barometer, 
is the same — for example, 30 inches. We may also draw lines 
connecting those places where the mean pressure is 29.9 inch- 



APPENDIX. 



255 




es, also 29. S inches, etc. ; or we may draw lines connecting all 
those places where the pressure is the same at any given day 
and hour, thus enabling us readily to follow the daily fluctna- 



256 APPENDIX. 

tions attending the progress of storms over the surface of the 
earth. 

The annexed figure shows the state of the barometer and 
the direction of the wind as observed near the centre of a vio- 
lent storm which prevailed in the neighborhood of New York 
February 16,1842. The small oval line shows the area within 



Portsmouth 




— .70 Inch. 



which the barometer sunk eight tenths of an inch below the 
mean, and the larger oval shows the area within which the 
barometer was depressed seven tenths of an inch. The long 
arrow represents the direction in which the storm advanced, 
while the short arrows show the observed direction of the wind 
at nearly forty different stations. 

Ex. 5. Lines of equal magnetic declination, dij), etc. We 
may draw upon a map of the earth curve lines connecting all 
those places where the declination of the magnetic needle is 
the same, or where the dip of the magnetic needle is the same, 
or the earth's magnetic intensity is the same. Such lines give 



APPENDIX. 



257 



a far more distinct idea of the distribution of magnetism over 
the earth's surface than could be furnished by any amount of 
numerical results exhibited in a tabular form. 

The annexed figure shows the lines of equal magnetic decli- 
nation for a portion of the United States for the year 1850. 




m 



TV T e perceive that the line of no declination passed through the 
centre of Lake Erie, and met the Atlantic near the middle of 
the coast of North Carolina. The line of 10 degrees west dec- 
lination passed near Montreal, and the line of 8 degrees east 
declination passed near St. Louis. These lines show a small 



258 APPENDIX. 

motion from year to year, and at present they all have a posi- 
tion westward of the positions represented on the map. 

The map also shows the line of 65° magnetic dip, of 70°, and 
of 75° dip. 

Ex. 6. How the principal phenomena of a storm may be 
represented. Winter storms in the United States are of great 
extent, sometimes exceeding 1000 miles in diameter. In order 
to represent the phenomena of such a storm, we require some 
suitable means of designating the area upon which rain or 
snow is falling ; we wish to denote the region around the mar- 
gin of the storm where clouds prevail without rain ; and we 
wish to represent the region of clear sky which encircles the 
storm on every side. We wish also to represent the depression 
of the barometer within the storm area ; also the state of the 
thermometer and the direction of the w^ind for each station of 
observation. The mode of accomplishing some of these ob- 
jects will be understood from the figure on the opposite page, 
which represents the principal phenomena of a violent storm 
which was experienced in the United States December 20, 
1836. The map represents the phenomena for 8 P.M. 

The deeply shaded portion in the middle of the figure rep- 
resents the area where rain or snow was falling; the lighter 
shade on the east and west margins of the rain represents the 
region where clouds prevailed without rain. Throughout the 
remaining portion of the United States, as far as the map ex- 
tends, clear sky prevailed. 

The dotted curve lines represent the state of the barometer. 
The inner curve shows the area where the barometer was de- 
pressed four tenths of an inch below the mean ; the next curve 
shows where the barometer was two tenths of an inch below 
the mean; the next curve shows the barometer at its mean 
height; while farther eastward the barometer stood two tenths 
of an inch and four tenths of an inch above the mean. 

The arrows show the directions of the wind as observed at a 
large number of stations. 

A similar map, constructed for 8 A.M., December 21, would 



APPENDIX. 



259 




show not only that the storm had traveled eastward, but that 
important changes had taken place within the storm area. 

This mode of representing the phenomena of a storm not 
merely compresses a vast amount of information within a small 
space, but it constitutes a powerful instrument of research, as 
it indicates a connection between the different classes of obser- 
vations which might entirely escape notice if the comparisons 
were limited to a collection of observations arranged in a tab- 
ular form. 



V. Relations of three independent variables. Since two co- 
ordinates are required to determine the position of a point on 
a plane, every point of a plane may be considered as corre- 
sponding to the known values of two of the variable elements. 
Take now three corresponding values of the three elements; 



2G0 APPENDIX. 

set off two of them as abscissa and ordinate on the given plane, 
and at the point thus determined erect a perpendicular whose 
length is proportional to the corresponding value of the third 
element. Proceed in the same manner with every three cor- 
responding values of the three variables. The extremities of 
all these perpendiculars will be situated upon a curved surface 
which represents the law connecting the three variable ele- 
ments. Suppose now a system of equidistant planes to be 
drawn parallel to the plane first assumed ; these planes will in- 
tersect the curved surface in curve lines whose form will indi- 
cate the undulations of that surface. Let these curves be now 
projected on the plane first assumed, and we shall have on a 
single plane a system of curve lines which give a precise idea 
of the changes of the third variable corresponding to any given 
change of the other two variables. 

Ex. Temperature at any hour and for any month. Let it 
be required to represent to the eye, by means of curve lines, 
the mean temperature of a given place for any hour of the day 
or any month of the yeaT. "We mark off on the axis of abscis- 
sas equal divisions to represent the months of the year, and on 
the axis of ordinates we set off, in like manner, twenty-four 
equal divisions to represent the hours of the day, and through 
these points of division we draw lines parallel to the co-ordinate 
axes. We are supposed to have a table, derived from observa- 
tion, which shows us the mean temperature of the given place 
for each hour and each month of the year. We now select 
any temperature — for example, 32° — and find the two hours of 
eacli month at which that temperature occurs. At the inter- 
section of the abscissa and ordinate corresponding to the given 
month and hour w T e place a point, and we do the same for each 
of the dates where the given temperature occurs. We join all 
these points by a continuous curve line, and we have a repre- 
sentation of the curve of 32°. In like manner we draw the 
curve of 30°, of 28°, etc., through the entire range of the ob- 
servations. The figure on the opposite page shows the results 
of a loner series of observations at New Haven. 



aitendix. 



2G1 




Noon 



October November December January February Marcli 



April 



May 



Such a figure shows at a glance the mean temperature cor- 
responding to any hour of either month of the year. If, for 
example, we desire to know the mean temperature of the month 
of January at 6 A.M., we find 6 A.M. on the left margin of the 
table, and follow along the corresponding horizontal line until 
we reach the middle of the month of January. The point 
falls nearly on the curve of 22°, which is therefore the temper- 
ature sought. In like manner we may find the temperature 
corresponding to any hour of any month of the year. 

The same figure shows the season of the year and the 
hour of the day when the lowest temperature occurs. It also 
shows, for any season of the year, the two hours which have 
the same temperature ; also, for any hour of the day, the two 
seasons of the year which have the same temperature. It also 
shows when the temperature changes most slowly, and when 
it changes most rapidly. 

In a similar manner we may construct a system of curve 
lines representing the relation between any three independent 
variables. 



THE END. 



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